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Ratios of a right triangle
A STORY
Two frames, same angle
Aarav and Meera stand in the maths lab with two wooden frames. Both are cut to the same sharp
angle. One frame is much taller than the other.
Aarav holds a plain wooden strip up beside the tall frame. Meera holds a strip just like it
beside the small frame, which stands on the table next to her.
“Same angle, different size,” Meera says. “My frame is a lot smaller than yours.”
“Suppose a triangle has a square corner,” Aarav says. “And I know one other angle and one side.”
Meera looks at her own small frame. “Then you can find the other two sides,” she says. “In a right
triangle, three ratios of the sides depend only on that angle. Not on how big it is.”
“Only on the angle?” Aarav asks. “Not the size at all?”
“Not the size at all,” Meera says. “A tiny right triangle and a huge one, same angle, same three
ratios.”
Could the same three ratios really hold for a right triangle ten times as tall?
Pick an acute angle inside a right triangle, and call it $A$. Three ratios of the
triangle’s sides depend on $A$ alone, never on the size of the triangle.
Draw a right triangle and mark angle $A$. Look at the side opposite $A$, and the hypotenuse.
Divide one by the other. Draw the triangle bigger or smaller. The ratio stays the same number,
for that one value of $A$. This ratio has a name, sine, written $\sin A$. A second
ratio, cosine, is $\cos A$. A third, tangent, is $\tan A$. Both are built from the triangle’s other
two sides.
The Baudhayana Sulba-sutra, an Indian altar-construction text from about 800 BCE, already used
this relation to lay out altars. Centuries later, Aryabhata’s Aryabhatiya (499 CE) tabulated sine
values like these, to compute planetary positions.
Five angles get exact values: $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$ and $90^\circ$.
One identity, $\sin^2 A + \cos^2 A = 1$, ties $\sin A$ and $\cos A$ together for every acute angle.
We prove both of these in full, not just state them.
Your turn: for a $3$-$4$-$5$ triangle, does $\sin^2 A + \cos^2 A$ come out to $1$? (Answer: yes.
$(3/5)^2 + (4/5)^2 = 9/25 + 16/25 = 1$.)
Picking the angle at vertex A names one side opposite and one adjacent; picking vertex C instead swaps both labels.
One triangle has three sides, and each of the six names picks two of them and writes one over the other.
Check yourself
In this chapter, the ratios $\sin A$, $\cos A$ and $\tan A$ for an acute angle $A$ of a right triangle depend on
the angle $A$ alone, never the size of the triangle
the size of the triangle, since a larger triangle gives larger ratios
which side the triangle-drawer chooses to call the hypotenuse
whether the sides are measured in centimetres or metres
Check your answer
✓ the angle $A$ alone, never the size of the triangle — (A) The ratios are fixed by the angle $A$ alone; the size of the triangle carrying that angle never matters.
the size of the triangle, since a larger triangle gives larger ratios — A larger similar triangle scales every side by the same factor, so the ratio of any two sides stays exactly the same.
which side the triangle-drawer chooses to call the hypotenuse — The hypotenuse is fixed as the side opposite the right angle — it is never a free choice of labelling.
whether the sides are measured in centimetres or metres — A ratio of two lengths in the same unit has no unit of its own — switching cm to m changes nothing about $\sin A$.
Suppose $\sin A$ really did change with the size of the right triangle carrying angle $A$. What would that break?
one angle would no longer give one fixed value of $\sin A$
nothing would break, since every right triangle still has a hypotenuse
Pythagoras’ theorem would stop holding for that triangle
the triangle would no longer be a right triangle
Check your answer
✓ one angle would no longer give one fixed value of $\sin A$ — (A) If the ratio changed with size, one angle would give many possible values, and no single value could ever be tabulated for it.
nothing would break, since every right triangle still has a hypotenuse — Every right triangle having a hypotenuse is not the issue — the problem is the ratio no longer being tied to the angle.
Pythagoras’ theorem would stop holding for that triangle — Pythagoras’ theorem relates the three sides of any right triangle and has nothing to do with whether a ratio is angle-only.
the triangle would no longer be a right triangle — The right angle is untouched by this scenario — only the constancy of the ratio is in question.
Two right triangles share the same acute angle $A$ but are different sizes. What can be said about $\tan A$ measured in each?
the larger triangle gives the larger value of $\tan A$
the two values differ by a fixed multiple, such as double or triple
the two triangles give exactly the same value of $\tan A$
no comparison is possible without knowing the exact side lengths
Check your answer
the larger triangle gives the larger value of $\tan A$ — $\tan A$ is opposite over adjacent; in a larger similar triangle both of those sides scale by the same factor, so the ratio itself does not grow.
the two values differ by a fixed multiple, such as double or triple — The values do not differ at all — the ratio is fixed by the angle alone, whatever the triangle’s size.
✓ the two triangles give exactly the same value of $\tan A$ — (C) $\tan A$ comes out exactly the same in both triangles, since the ratio is fixed by the angle alone.
no comparison is possible without knowing the exact side lengths — No side length is needed — the angle alone fixes the ratio, in either triangle.
You already know the shapes and the numbers behind this. Now put them together, one ratio at a time. Try each check below.
Pythagoras’ theorem (Class 8, Ganita Prakash Part 2, page 44). Here: the identity $\sin^2 A + \cos^2 A = 1$ needs $p^2 + b^2 = h^2$. Check: $3^2 + 4^2 = 5^2$, since $9 + 16 = 25$.
A ratio in simplest form (Class 8, Ganita Prakash Part 1, page 161). Here: sides $4k$ and $3k$ use a ratio already in this form. Check: $4 : 3$ has no common factor, so it is already in simplest form.
Undoing a square (Class 8, Ganita Prakash Part 1, page 8). Here: finding a third side, $\sqrt{25 k^2} = 5 k$, undoes the square this way. Check: $\sqrt{25} = 5$, since $5^2 = 25$.
Irrational numbers (Class 9, Ganita Manjari, page 53). Here: $\sin 60^\circ = \sqrt{3}/2$ has no exact fraction. Check: $\sqrt{2}$ squared is exactly $2$, though $\sqrt{2}$ itself is not a fraction.
The difference of squares (Class 9, Ganita Manjari, page 77). Here: a proof turns $(1 - \sin A)(1 + \sin A)$ into $1 - \sin^2 A$ this way. Check: $(5 - 2) \cdot (5 + 2) = 5^2 - 2^2$, both equal to $21$.
Squaring a number (Class 8, Ganita Prakash Part 1, page 21). Here: $\sin^2 A$ means $\sin A$ raised to the power $2$, the notation used here. Check: $3$ raised to the power $2$ is $9$.
If any of these felt new, read the page named before going on.
Pythagoras is a statement about areas, and every identity in this chapter is that same statement divided through.
Before naming any ratio, we name the sides. The hypotenuse is the side opposite the
right angle. It is the longest side, and its name never changes with $A$.
The other two sides are named relative to $A$ itself. The opposite side faces $A$ directly. The
adjacent side also touches $A$, but it is not the hypotenuse.
We build three ratios from these three sides. $\sin A$ is opposite over hypotenuse. $\cos A$ is
adjacent over hypotenuse. $\tan A$ is opposite over adjacent.
Pick a different acute angle in the same triangle, and the two names swap. Name the three sides
correctly for the angle in question first: that is step one of every problem.
Your turn: in a right triangle, which side is adjacent to angle $A$? (Answer: the side that
touches $A$, but is not the hypotenuse.)
Check yourself
For an acute angle $A$ in a right triangle, $\sin A$ is defined as
the side adjacent to $A$ over the hypotenuse
the side opposite $A$ over the hypotenuse
the side opposite $A$ over the side adjacent to $A$
the hypotenuse over the side opposite $A$
Check your answer
the side adjacent to $A$ over the hypotenuse — Adjacent over hypotenuse is $\cos A$, not $\sin A$ — the numerator here is the side facing $A$, not the side beside it.
✓ the side opposite $A$ over the hypotenuse — (B) $\sin A$ is the side opposite $A$ divided by the hypotenuse, by definition.
the side opposite $A$ over the side adjacent to $A$ — Opposite over adjacent is $\tan A$; $\sin A$ always has the hypotenuse in the denominator.
the hypotenuse over the side opposite $A$ — Hypotenuse over opposite is the reciprocal ratio $\text{cosec} A$, not $\sin A$ itself.
Why does $\tan A = (\sin A)/(\cos A)$ hold for every acute angle $A$?
because the hypotenuse cancels when $\sin A$ is divided by $\cos A$
because $\tan A$ is defined that way by a separate rule unrelated to $\sin A$ and $\cos A$
because the identity is only approximately true for small angles
because $\cos A$ is always larger than $\sin A$
Check your answer
✓ because the hypotenuse cancels when $\sin A$ is divided by $\cos A$ — (A) $\sin A$ and $\cos A$ both carry the hypotenuse as their denominator, so dividing one by the other cancels it, leaving opposite over adjacent — exactly $\tan A$.
because $\tan A$ is defined that way by a separate rule unrelated to $\sin A$ and $\cos A$ — $\tan A$ is opposite over adjacent by definition, and dividing $\sin A$ by $\cos A$ reaches that same ratio because the hypotenuse cancels — it is not a separate, unrelated rule.
because the identity is only approximately true for small angles — The identity is exact for every acute angle, not an approximation restricted to small ones.
because $\cos A$ is always larger than $\sin A$ — Whether $\cos A$ exceeds $\sin A$ has nothing to do with why the quotient identity holds — it holds because both ratios share the hypotenuse as a denominator.
Three more ratios are just the first three turned upside down. $\text{cosec} A = 1/(\sin A)$.
$\sec A = 1/(\cos A)$. $\cot A = 1/(\tan A)$. Each one is the reciprocal of a ratio we already named.
$\tan A$ also has a second life, as a quotient. $\tan A = (\sin A)/(\cos A)$. Its reciprocal follows
the same pattern: $\cot A = (\cos A)/(\sin A)$.
The four newer names are bookkeeping on top of the first two, not four fresh ideas. If a ratio
will not come to mind directly, rewrite it through $\sin A$ and $\cos A$ first. You will usually
find this the fastest way in.
Your turn: what is $\sec A$ in terms of $\cos A$? (Answer: $\sec A = 1/(\cos A)$.)
Any exercise asking for cosecant, secant or cotangent wants the reciprocal of a ratio you already found, not a fresh triangle measurement.
Check yourself
In a right triangle, $\sin A = 3/5$ and $\cos A = 4/5$. Using $\cot A = (\cos A)/(\sin A)$, what is $\cot A$?
$3/4$
$4/3$
$5/3$
$5/4$
Check your answer
$3/4$ — $3/4$ is $\tan A$ itself; $\cot A$ is its reciprocal, $4/3$.
✓ $4/3$ — (B) $\cot A = (\cos A)/(\sin A) = (4/5)/(3/5) = 4/3$.
$5/3$ — $5/3$ is $\text{cosec} A$ (hypotenuse over opposite); $\cot A$ is $\cos A$ over $\sin A$.
$5/4$ — $5/4$ is $\sec A$ (hypotenuse over adjacent); $\cot A$ is $\cos A$ over $\sin A$.
$\sec A$ is defined as the reciprocal of
$\sin A$
$\tan A$
$\cot A$
$\cos A$
Check your answer
$\sin A$ — The reciprocal of $\sin A$ is $\text{cosec} A$; $\sec A$ is the reciprocal of $\cos A$.
$\tan A$ — The reciprocal of $\tan A$ is $\cot A$; $\sec A$ reciprocates $\cos A$, not $\tan A$.
$\cot A$ — $\cot A$ is the reciprocal of $\tan A$; $\sec A$ reciprocates $\cos A$.
✓ $\cos A$ — (D) $\sec A = 1/(\cos A)$, the reciprocal of $\cos A$.
This chapter says every one of the six ratios reduces to just $\sin A$ and $\cos A$. Why is that true even for $\text{cosec} A$ and $\sec A$?
because a calculator always converts them to $\sin A$ and $\cos A$ first
because $\text{cosec} A$ and $\sec A$ do not actually depend on the angle $A$
because both are defined as reciprocals of $\sin A$, $\cos A$
because the hypotenuse cancels out of every ratio
Check your answer
because a calculator always converts them to $\sin A$ and $\cos A$ first — No calculator step is involved — $\text{cosec} A$ and $\sec A$ reduce to $\sin A$ and $\cos A$ purely by their own definitions as reciprocals.
because $\text{cosec} A$ and $\sec A$ do not actually depend on the angle $A$ — $\text{cosec} A$ and $\sec A$ depend on $A$ exactly as $\sin A$ and $\cos A$ do — they are simply their reciprocals.
✓ because both are defined as reciprocals of $\sin A$, $\cos A$ — (C) $\text{cosec} A = 1/(\sin A)$ and $\sec A = 1/(\cos A)$ by definition, so both are already written in terms of $\sin A$ and $\cos A$.
because the hypotenuse cancels out of every ratio — The hypotenuse does not cancel out of $\text{cosec} A$ or $\sec A$ — it stays in the ratio; the reduction to $\sin A$/$\cos A$ comes from the reciprocal definitions, not cancellation.
Given $\sin A = 8/17$ and $\cos A = 15/17$, use $\cot A = (\cos A)/(\sin A)$ to find $\cot A$.
$8/15$
$17/8$
$17/15$
$15/8$
Check your answer
$8/15$ — $8/15$ is $\tan A$; $\cot A$ is its reciprocal, $15/8$.
$17/8$ — $17/8$ is $\text{cosec} A$; $\cot A$ is $\cos A$ over $\sin A$, $15/8$.
$17/15$ — $17/15$ is $\sec A$; $\cot A$ is $\cos A$ over $\sin A$, $15/8$.
We use one Class 8 result throughout. For a right triangle with legs $a$ and $b$ and
hypotenuse $c$, $c^2 = a^2 + b^2$. We do not prove it again here.
Identifying the hypotenuse correctly matters more than the formula itself. It is the side opposite
the right angle, and it is always the longest of the three sides. Check this before substituting
into the relation.
This same relation proves $\sin^2 A + \cos^2 A = 1$ soon, without being re-derived.
Your turn: a right triangle has legs $6$ and $8$. What is the hypotenuse? (Answer:
$c = \sqrt{36 + 64} = 10$.)
Check yourself
In a right triangle with legs $a$ and $b$ and hypotenuse $c$, Pythagoras’ theorem states
$c = a^2 + b^2$
$c^2 = a^2 - b^2$
$a^2 = b^2 + c^2$
$c^2 = a^2 + b^2$
Check your answer
$c = a^2 + b^2$ — The hypotenuse itself is squared too: it is $c^2$, not $c$, that equals $a^2+b^2$.
$c^2 = a^2 - b^2$ — The two legs’ squares are added, never subtracted, to give the hypotenuse’s square.
$a^2 = b^2 + c^2$ — It is the hypotenuse’s square that equals the sum of the two legs’ squares, not one leg’s square.
✓ $c^2 = a^2 + b^2$ — (D) Pythagoras’ theorem states $c^2 = a^2+b^2$, the hypotenuse’s square equal to the sum of the legs’ squares.
This chapter proves $\sin^2 A + \cos^2 A = 1$ using Pythagoras’ theorem. What role does the theorem play in that proof?
it defines what $\sin A$ and $\cos A$ mean in the first place
it proves that the triangle has a right angle
it shows the triangle’s angles add to $180^\circ$
it supplies the side relation the proof relies on
Check your answer
it defines what $\sin A$ and $\cos A$ mean in the first place — $\sin A$ and $\cos A$ are defined earlier, from opposite/adjacent over hypotenuse; Pythagoras’ theorem is the separate side-relation the proof then uses.
it proves that the triangle has a right angle — The right angle is assumed to start with; Pythagoras’ theorem is a consequence of it, used here to relate the sides.
it shows the triangle’s angles add to $180^\circ$ — The angle-sum fact plays no role here — the proof only needs the side relation $p^2+b^2=h^2$.
✓ it supplies the side relation the proof relies on — (D) Pythagoras’ theorem supplies the side relation $p^2+b^2=h^2$ that the entire proof of the identity is built on.
A right triangle has legs $6$ and $8$. Using Pythagoras’ theorem, its hypotenuse is
$14$
$10$
$48$
$\sqrt{28}$
Check your answer
$14$ — Adding $6+8=14$ skips the theorem entirely — the legs must be squared, added, then rooted: $\sqrt{36+64}=10$.
Every ratio so far rests on something not yet proved. $\sin A$, $\cos A$ and $\tan A$
depend only on angle $A$, never on the triangle’s size. Let us prove it now,
using similarity from Chapter 6.
*THEOREM. Fix an acute angle $A$. Every right triangle carrying $A$ gives the same $\sin A$,
$\cos A$ and $\tan A$. The size of the triangle makes no difference.*
We show it directly. Any two right triangles sharing angle $A$ turn out similar to each other.
That similarity then forces every ratio to match between them. The numbered proof below carries
each step with its own reason.
Your turn: a $3$-$4$-$5$ triangle and a $6$-$8$-$10$ triangle both carry angle $A$. Should $\sin A$
come out the same for both? (Answer: yes, by the proof below.)
Proof
Given. We take two right triangles that both carry the same acute angle $A$. In the first, the side opposite $A$ is $p_1$, the side adjacent to $A$ is $b_1$, and the hypotenuse is $h_1$. In the second, the matching sides are $p_2$, $b_2$ and $h_2$.
To prove. For an acute angle $A$, the ratios $\sin A$, $\cos A$ and $\tan A$ are the same for every right triangle that carries $A$.
let two right triangles both carry the same acute angle $A$ In the first triangle, we call the side opposite $A$ as $p_1$, the side adjacent to $A$ as $b_1$, and the hypotenuse as $h_1$. We give the second triangle’s matching sides the labels $p_2$, $b_2$ and $h_2$. The two triangles may be different sizes.
the two triangles are similar Both triangles have a right angle, and both have angle $A$. By AA similarity (Chapter 6), the triangles are similar.
$p_1/p_2 = b_1/b_2 = h_1/h_2$ Similar triangles have proportional corresponding sides.
$p_1/h_1 = p_2/h_2$ We rearrange $p_1/p_2 = h_1/h_2$ by cross-multiplying the proportion.
$\sin A$ measured in the first triangle equals $\sin A$ measured in the second The left side of the previous step is $\sin A$ in the first triangle. The right side is $\sin A$ in the second. This follows since $\sin A$ is opposite over hypotenuse, by definition.
$\cos A$ and $\tan A$ agree in both triangles too We apply the same rearrangement to $b_1/b_2 = h_1/h_2$ and to $p_1/p_2 = b_1/b_2$.
every right triangle carrying angle $A$ gives the same three ratios The two triangles we compared were any two right triangles sharing $A$. So the argument holds for every such pair. Each ratio depends on the angle alone, never on the triangle’s size.
■
Because this holds for any two triangle sizes sharing an angle, a numeric example can pick the easiest size to compute with.
A SCHOLAR INDIA REMEMBERS
Draw a small right triangle with a $30^\circ$ angle. Then draw a big one. Will $\sin 30^\circ$ come out the same? Count it. In the small one, the opposite side is $1$ and the hypotenuse is $2$. In the big one, they are $3$ and $6$. Both give $1/2$. The ratio depends on the angle, not on the size.
Aryabhata · the hoopoe
Check yourself
The proof that $\sin A$, $\cos A$ and $\tan A$ depend only on the angle $A$ rests on
the SSS congruence criterion, applied to two right triangles sharing angle $A$
the AA similarity criterion, applied to two right triangles sharing angle $A$
measuring the ratios in many different-sized triangles and averaging the results
the fact that every right triangle has one angle equal to $90^\circ$
Check your answer
the SSS congruence criterion, applied to two right triangles sharing angle $A$ — Congruence would force the triangles to be the same size; the proof only needs them similar, which AA similarity gives.
✓ the AA similarity criterion, applied to two right triangles sharing angle $A$ — (B) The proof uses AA similarity: two right triangles sharing angle $A$ are similar, and similar triangles have proportional sides.
measuring the ratios in many different-sized triangles and averaging the results — The claim is proved exactly, from similarity — it is not an empirical average over many measured triangles.
the fact that every right triangle has one angle equal to $90^\circ$ — Every right triangle sharing a right angle is not enough on its own — it is sharing angle $A$ as well that makes the two triangles similar.
In the proof, two right triangles share the same acute angle $A$ but have sides $p_1,b_1,h_1$ and $p_2,b_2,h_2$. Why does AA similarity apply to them?
both triangles have all three sides equal in length
both triangles have the same area
both triangles were drawn using the same ruler
both share a right angle and angle $A$
Check your answer
both triangles have all three sides equal in length — The proof never assumes equal sides — that would make the triangles congruent, not merely similar, and is far stronger than needed.
both triangles have the same area — Area is never compared in this proof — AA similarity needs only two matching angles.
both triangles were drawn using the same ruler — What tool drew the triangles is irrelevant — similarity is a statement about the angles, not the instrument used.
✓ both share a right angle and angle $A$ — (D) Two matching angles, the right angle and angle $A$, are already enough to make the two triangles similar by AA.
The proof uses $p_1/p_2 = h_1/h_2$ from similar triangles and rearranges it to $p_1/h_1 = p_2/h_2$. What does this rearranged line show?
that the two triangles have the same hypotenuse
that $p_1$ equals $p_2$
that $\sin A$ matches in both triangles
that the two triangles are congruent
Check your answer
that the two triangles have the same hypotenuse — The two hypotenuses $h_1$ and $h_2$ need not be equal at all — only the ratio $p/h$ agrees between the triangles, not the raw lengths.
that $p_1$ equals $p_2$ — The two opposite sides $p_1$ and $p_2$ can differ freely — only their ratio to the matching hypotenuse agrees.
✓ that $\sin A$ matches in both triangles — (C) $p_1/h_1$ is $\sin A$ in the first triangle and $p_2/h_2$ is $\sin A$ in the second, so this line shows the two agree.
that the two triangles are congruent — One matching ratio shows the triangles are similar, which the proof already knew — it says nothing about them being the same size.
A student measures $\sin 40^\circ$ using a small right triangle and gets $0.643$. A classmate measures $\sin 40^\circ$ using a much larger right triangle with the same angle. What should the classmate get?
a larger number than $0.643$, since the sides are all larger
a smaller number than $0.643$, since a bigger triangle spreads the angle out more
it cannot be predicted without knowing the larger triangle’s exact side lengths
$0.643$ again, since the ratio does not depend on the triangle’s size
Check your answer
a larger number than $0.643$, since the sides are all larger — Every side in the larger triangle scales by the same factor, so the ratio of two sides is unchanged — the answer stays $0.643$.
a smaller number than $0.643$, since a bigger triangle spreads the angle out more — The triangle’s size does not change how spread out the angle looks — angle $A$ is exactly $40^\circ$ in both triangles, unaffected by scale.
it cannot be predicted without knowing the larger triangle’s exact side lengths — No measurement of the larger triangle is needed — the proof guarantees the same value from the angle alone.
✓ $0.643$ again, since the ratio does not depend on the triangle’s size — (D) The proof guarantees the same value, $0.643$, whatever the size of the triangle carrying the angle.
Chapter 6 proved that similar triangles have proportional sides before this chapter existed. Why did that have to come first?
because Chapter 6 introduced the value of $\pi$, needed later in this chapter
because Chapter 6 proved Pythagoras’ theorem for the first time
because Chapter 6 is simply the previous chapter, and chapters are always used in order
because the proof directly uses Chapter 6’s similarity result
Check your answer
because Chapter 6 introduced the value of $\pi$, needed later in this chapter — $\pi$ plays no part in this chapter — the link to Chapter 6 is through similarity, not circles.
because Chapter 6 proved Pythagoras’ theorem for the first time — Pythagoras’ theorem is a Class 8 result, stated here without proof — Chapter 6’s contribution to this chapter is similarity, not Pythagoras.
because Chapter 6 is simply the previous chapter, and chapters are always used in order — Chapters are not needed just because they come earlier — Chapter 6 is needed here specifically because its similarity result is the tool this proof runs on.
✓ because the proof directly uses Chapter 6’s similarity result — (D) Chapter 6’s similarity result is exactly the tool this chapter’s proof runs on, so it had to come first.
The hypotenuse is the longest side of a right triangle. That one fact bounds every ratio we
have defined so far.
For an acute angle $A$, $\sin A$ and $\cos A$ are each a shorter side over the hypotenuse. So both
are positive, and never more than $1$. Their reciprocals flip that bound: $\text{cosec} A$ and $\sec A$
are always at least $1$. $\tan A$ compares the two legs directly, with no hypotenuse involved, so
it has no upper bound at all.
A claimed value like $\sin A = 4/3$ is impossible on sight. Check a ratio you compute against this
bound: it catches an arithmetic slip early.
Your turn: is $\sec A = 2/3$ possible for an acute angle? (Answer: no. $\sec A$ can never be less
than $1$.)
This same argument holds on any right triangle: a leg divided by the hypotenuse can never reach or pass 1.
A SCHOLAR INDIA REMEMBERS
Put a bound on it before you work it out. In a right triangle the hypotenuse is the longest side. So opposite over hypotenuse is less than $1$. That means $\sin A$ can never be $1.2$. If you get $1.2$, go back and look for the slip.
Aryabhata · the hoopoe
Check yourself
For an acute angle $A$, the value of $\sin A$
can be greater than $1$ for a large enough angle
can be negative for some acute angles
equals exactly $1$ for every acute angle
is always positive and never more than $1$
Check your answer
can be greater than $1$ for a large enough angle — The hypotenuse is always the longest side of a right triangle, so opposite over hypotenuse can never exceed $1$.
can be negative for some acute angles — For an acute angle, both the opposite side and the hypotenuse are positive lengths, so $\sin A$ is always positive.
equals exactly $1$ for every acute angle — $\sin A$ reaches $1$ only in the limit as $A$ approaches $90^\circ$ — for acute $A$ it stays strictly below $1$.
✓ is always positive and never more than $1$ — (D) $\sin A$ is always positive and never exceeds $1$, since the opposite side can never exceed the hypotenuse.
Why can $\sec A$ never be less than $1$ for an acute angle $A$?
because $\sec A$ is measured in different units from $\cos A$
because $\sec A=1/(\cos A)$ and $\cos A\leq 1$
because the hypotenuse is always an integer
because $\sec A$ is defined as the adjacent side over the opposite side
Check your answer
because $\sec A$ is measured in different units from $\cos A$ — $\sec A$ and $\cos A$ are both unit-free ratios; the bound comes from the reciprocal relationship, not from units.
✓ because $\sec A=1/(\cos A)$ and $\cos A\leq 1$ — (B) $\sec A$ is the reciprocal of $\cos A$; since $\cos A$ never exceeds $1$, its reciprocal $\sec A$ never falls below $1$.
because the hypotenuse is always an integer — The hypotenuse need not be an integer at all — the bound on $\sec A$ follows purely from $\cos A \leq 1$.
because $\sec A$ is defined as the adjacent side over the opposite side — $\sec A$ is the reciprocal of $\cos A$ (hypotenuse over adjacent), not adjacent over opposite — that ratio is $\cot A$.
A student claims to have measured $\cos A = 1.2$ for some acute angle $A$ in a right triangle. What does this claim reveal?
the angle $A$ must be close to $90^\circ$
the triangle must have an unusually long hypotenuse
the student measured $\sec A$ by mistake instead of $\cos A$
the measurement is impossible; $\cos A$ never exceeds $1$
Check your answer
the angle $A$ must be close to $90^\circ$ — No acute angle gives $\cos A$ above $1$ — the value itself is impossible, not just unusually large.
the triangle must have an unusually long hypotenuse — However long the hypotenuse is, $\cos A$ stays adjacent over hypotenuse, which cannot exceed $1$ in any right triangle.
the student measured $\sec A$ by mistake instead of $\cos A$ — Nothing in the claim suggests $\sec A$ was measured instead — the direct conclusion is simply that the stated value is impossible for $\cos A$.
✓ the measurement is impossible; $\cos A$ never exceeds $1$ — (D) No acute angle can give $\cos A = 1.2$ — the claim is impossible, since $\cos A$ never exceeds $1$.
This same two-sides-then-Pythagoras move is the one method behind any item that gives a single ratio and asks for the rest.
Given just one ratio of an acute angle $A$, we can recover every other ratio of that same
angle. No triangle needs to be measured.
Write the given ratio as a fraction in lowest terms. Scale both the numerator and the denominator
by the same positive number $k$. This fixes two of the triangle’s three sides. Pythagoras’ theorem
then supplies the third side, and every ratio can be read directly off the three.
Scaling only one side breaks the right angle the whole recipe depends on. Given $\tan A = 5/12$,
this fixes opposite $= 5 k$ and adjacent $= 12 k$. The hypotenuse is not yet found.
Your turn: given $\sin A = 5/13$, what two sides does this fix, before finding the third? (Answer:
opposite $= 5 k$, hypotenuse $= 13 k$.)
Check yourself
Given that $\sin A = 2/3$ for an acute angle $A$, the standard recipe to find every other ratio starts by
writing the opposite side as $2$ and the adjacent side as $3$
measuring the actual triangle with a ruler to find the sides
assuming the triangle is equilateral
opposite $=2k$, hypotenuse $=3k$, for some $k>0$
Check your answer
writing the opposite side as $2$ and the adjacent side as $3$ — $\sin A$’s denominator is the hypotenuse, not the adjacent side — the recipe scales opposite $=2k$, hypotenuse $=3k$.
measuring the actual triangle with a ruler to find the sides — No physical triangle is given to measure — the recipe builds the sides algebraically from the ratio itself.
assuming the triangle is equilateral — An equilateral triangle has no right angle at all — nothing about $\sin A = 2/3$ assumes this shape.
✓ opposite $=2k$, hypotenuse $=3k$, for some $k>0$ — (D) The recipe scales the given fraction: opposite $=2k$, hypotenuse $=3k$, for some positive $k$.
In the ratio-to-triangle recipe, after writing two sides as $2k$ and $3k$, why is Pythagoras’ theorem used next?
to check whether the angle $A$ is really acute
to convert the ratio into a decimal number
to find the triangle’s third, still-missing side
because the two given sides might not actually form a right triangle
Check your answer
to check whether the angle $A$ is really acute — Pythagoras’ theorem finds a missing side length — whether $A$ is acute is already given, not something the theorem checks.
to convert the ratio into a decimal number — The theorem finds the missing side length; no decimal conversion of the ratio is involved.
✓ to find the triangle’s third, still-missing side — (C) The given ratio only fixes two sides; Pythagoras’ theorem is needed to find the third.
because the two given sides might not actually form a right triangle — The triangle is a right triangle by the setup — the theorem is used to find the missing side, not to test the right angle.
Given $\cos A = 5/13$ for an acute angle $A$, the recipe gives adjacent $=5k$, hypotenuse $=13k$, opposite $=12k$. What is $\sin A$?
$5/12$
$12/13$
$13/12$
$5/13$
Check your answer
$5/12$ — $5/12$ divides adjacent by opposite, which is neither $\sin A$ nor a ratio this recipe asks for; $\sin A$ is opposite over hypotenuse, $12/13$.
✓ $12/13$ — (B) $\sin A =$ opposite over hypotenuse $= 12k/13k = 12/13$.
$13/12$ — $13/12$ is $\text{cosec} A$, the reciprocal of $\sin A$; $\sin A$ itself is $12/13$.
$5/13$ — $5/13$ is the $\cos A$ value already given in the question; $\sin A$ still needs computing from the missing opposite side, $12/13$.
Neither triangle here can be drawn at exactly 0 degrees or 90 degrees, because no right triangle can carry that angle.
The ratios at $0^\circ$ and $90^\circ$ are not read off an actual triangle. No triangle
can carry an angle of exactly $0^\circ$ or $90^\circ$ alongside its own right angle. Instead, we
watch what happens as $A$ approaches them.
As angle $A$ shrinks toward $0^\circ$, the side opposite $A$ shrinks toward nothing. The
adjacent side grows to nearly the full hypotenuse. This gives $\sin 0^\circ = 0$ and
$\cos 0^\circ = 1$. Run the same reasoning as $A$ grows toward $90^\circ$.
$\sin 90^\circ = 1$ and $\cos 90^\circ = 0$ follow the same way.
*$\tan 90^\circ$ is not defined, because its definition divides by $\cos 90^\circ = 0$.
Dividing by zero is never allowed.*
From $\sin 0^\circ = 0$ and $\cos 0^\circ = 1$, $\tan 0^\circ = 0$ follows directly.
$\sec 90^\circ$, $\text{cosec} 0^\circ$ and $\cot 0^\circ$ fail for the same reason.
Your turn: why is $\text{cosec} 0^\circ$ undefined? (Answer: $\text{cosec} 0^\circ = 1/(\sin 0^\circ) = 1/0$, and division by zero is not allowed.)
Check yourself
As angle $A$ shrinks toward $0^\circ$ in a right triangle, the opposite side shrinks to nothing while the adjacent side nearly equals the hypotenuse. This motivates
$\sin 0^\circ = 1$ and $\cos 0^\circ = 0$
$\sin 0^\circ = 0$ and $\cos 0^\circ = 0$
$\sin 0^\circ = 0$ and $\cos 0^\circ = 1$
$\tan 0^\circ = 1$
Check your answer
$\sin 0^\circ = 1$ and $\cos 0^\circ = 0$ — It is the opposite side that shrinks to nothing at $0^\circ$, giving $\sin 0^\circ = 0$, not $1$; $\cos 0^\circ = 1$.
$\sin 0^\circ = 0$ and $\cos 0^\circ = 0$ — The adjacent side nearly equals the hypotenuse as $A$ shrinks, so $\cos A$ approaches $1$, not $0$.
✓ $\sin 0^\circ = 0$ and $\cos 0^\circ = 1$ — (C) A vanishing opposite side gives $\sin 0^\circ = 0$, and an adjacent side nearly equal to the hypotenuse gives $\cos 0^\circ = 1$.
Why is $\tan 90^\circ$ not defined, while $\tan 0^\circ$ equals $0$?
because no right triangle can ever contain a $90^\circ$ angle
because $\tan A$ is only defined for angles below $45^\circ$
because dividing by $\cos 90^\circ=0$ is not allowed
because $\sin 90^\circ$ is also not defined
Check your answer
because no right triangle can ever contain a $90^\circ$ angle — The right angle of the triangle is a separate, fixed $90^\circ$ — this is about angle $A$ approaching $90^\circ$, at which point $\cos A$ hits $0$ and division breaks.
because $\tan A$ is only defined for angles below $45^\circ$ — $\tan A$ is defined for every acute angle up to (but not including) $90^\circ$ — there is no cutoff at $45^\circ$.
✓ because dividing by $\cos 90^\circ=0$ is not allowed — (C) $\tan A$ is $\sin A$ over $\cos A$; at $90^\circ$ the denominator $\cos 90^\circ = 0$, making the division impossible.
because $\sin 90^\circ$ is also not defined — $\sin 90^\circ = 1$ is perfectly defined — it is dividing by $\cos 90^\circ = 0$ that breaks $\tan 90^\circ$, not sin’s own value.
A right triangle’s acute angle $A$ is made smaller and smaller while the hypotenuse stays fixed. What happens to $\cot A$?
$\cot A$ shrinks toward $0$, following the opposite side
$\cot A$ stays fixed at $1$, since the hypotenuse does not change
$\cot A$ grows without bound as $\sin A$ shrinks toward $0$
$\cot A$ becomes negative
Check your answer
$\cot A$ shrinks toward $0$, following the opposite side — It is $\sin A$ that shrinks toward $0$; $\cot A$ is $\cos A$ divided by that shrinking number, so it grows rather than shrinks.
$\cot A$ stays fixed at $1$, since the hypotenuse does not change — The hypotenuse staying fixed does not fix the ratio — as $A$ shrinks, $\sin A$ falls toward $0$ while $\cos A$ rises toward $1$, so their quotient $\cot A$ grows.
✓ $\cot A$ grows without bound as $\sin A$ shrinks toward $0$ — (C) As $A$ shrinks, $\sin A$ falls toward $0$ while $\cos A$ rises toward $1$, so $\cot A = (\cos A)/(\sin A)$ grows without bound.
$\cot A$ becomes negative — For every acute angle, both $\sin A$ and $\cos A$ stay positive, so $\cot A$ stays positive too — it grows large, never negative.
Every 30 degree and 60 degree value in the standard-angle table comes from this one triangle, so a forgotten entry can be rebuilt.
Five angles carry exact values worth memorising: $0^\circ$, $30^\circ$, $45^\circ$,
$60^\circ$ and $90^\circ$.
Sine runs $0$, then $1/2$, then $1/\sqrt{2}$, then $\sqrt{3}/2$, then $1$. Cosine runs the same five
numbers in reverse: $1$, then $\sqrt{3}/2$, then $1/\sqrt{2}$, then $1/2$, then $0$. $\tan A$ is
$\sin A$ divided by $\cos A$ at each angle: $0$, then $1/\sqrt{3}$, then $1$, then $\sqrt{3}$.
$\tan A$ has no value at $90^\circ$, since $\cos A$ is $0$ there. Every reciprocal ratio follows
by inverting the matching entry.
The two rows run through the same five numbers in opposite order. We do not prove why here,
though it is worth noticing. Working $\tan A$ out from $\sin A$ and $\cos A$ is usually faster than
memorising a third list. A right triangle with two equal legs gives $45^\circ$. Half an
equilateral triangle gives $30^\circ$ and $60^\circ$ instead.
Your turn: what is $\cos 60^\circ$? (Answer: $1/2$, reading the cosine row at $60^\circ$.)
The 45 degree column of the standard-angle table comes from exactly this triangle, with no other construction needed.
A SCHOLAR INDIA REMEMBERS
Aryabhata’s mathematics rested on geometric reasoning. A result was accepted because it could be worked out, not because someone important said so. You can work out this table the same way. Take an equilateral triangle with sides $2$. Cut it in half. The half has angles $30^\circ$, $60^\circ$ and $90^\circ$. The side opposite $30^\circ$ is $1$, and the hypotenuse is $2$. So $\sin 30^\circ = 1/2$.
Aryabhata · the hoopoe
Check yourself
At $A = 45^\circ$, the values of $\sin A$ and $\cos A$ are
both equal to $1/\sqrt{2}$
$\sin 45^\circ = 1/2$ and $\cos 45^\circ = \sqrt{3}/2$
$\sin 45^\circ = \sqrt{3}/2$ and $\cos 45^\circ = 1/2$
$\sin 45^\circ = 1$ and $\cos 45^\circ = 0$
Check your answer
✓ both equal to $1/\sqrt{2}$ — (A) At $45^\circ$, $\sin A = \cos A = 1/\sqrt{2}$.
$\sin 45^\circ = 1/2$ and $\cos 45^\circ = \sqrt{3}/2$ — $1/2$ and $\sqrt{3}/2$ are the $30^\circ$ values; at $45^\circ$ both $\sin$ and $\cos$ equal $1/\sqrt{2}$.
$\sin 45^\circ = \sqrt{3}/2$ and $\cos 45^\circ = 1/2$ — $\sqrt{3}/2$ and $1/2$ are the $60^\circ$ values; at $45^\circ$ both $\sin$ and $\cos$ equal $1/\sqrt{2}$.
$\sin 45^\circ = 1$ and $\cos 45^\circ = 0$ — $1$ and $0$ are the $90^\circ$ values; at $45^\circ$ both $\sin$ and $\cos$ equal $1/\sqrt{2}$.
The standard-angle table lists $\tan$ as not defined at $90^\circ$ but gives an exact value at every other listed angle. Why?
because no triangle can have an angle of exactly $90^\circ$
because $\sin 90^\circ$ is undefined, and $\tan$ needs $\sin$
because $\cos 90^\circ=0$ makes the division impossible
because the table only lists five angles by convention, with no deeper reason
Check your answer
because no triangle can have an angle of exactly $90^\circ$ — This is about angle $A$, not the triangle’s own right angle — $\cos 90^\circ = 0$ is what breaks the division, not the triangle’s shape.
because $\sin 90^\circ$ is undefined, and $\tan$ needs $\sin$ — $\sin 90^\circ = 1$ is perfectly defined; it is $\cos 90^\circ = 0$ in the denominator that breaks $\tan 90^\circ$.
✓ because $\cos 90^\circ=0$ makes the division impossible — (C) $\tan A$ is $\sin A$ over $\cos A$, and $\cos 90^\circ = 0$ makes that particular division impossible.
because the table only lists five angles by convention, with no deeper reason — The gap is not a convention — dividing by $\cos 90^\circ = 0$ is genuinely impossible, wherever it appears.
Using the standard-angle table, evaluate $\sin 30^\circ \cdot \cos 60^\circ + \cos 30^\circ \cdot \sin 60^\circ$.
$1/2$ — $\sin 30^\circ \cdot \cos 60^\circ = 1/2 \cdot 1/2 = 1/4$ is only the first term; the second term $\cos 30^\circ \cdot \sin 60^\circ = 3/4$ must be added too, giving $1$.
$0$ — The expression adds the two products, giving $1$; subtracting them instead gives $1/2$, not what the stem asks for.
$\sqrt{3}/2$ — $\sqrt{3}/2$ is a single table entry ($\cos 30^\circ$ or $\sin 60^\circ$), not the value of the full sum, which is $1$.
Because no two of these five heights repeat, one ratio value always points back to exactly one angle, never two.
Acute angle $A$ runs from $0^\circ$ to $90^\circ$. As it does, $\sin A$ climbs from $0$
to $1$. $\cos A$ falls from $1$ to $0$ over the same range. Neither ratio ever doubles back.
*Two right triangles giving the same value of $\sin A$ must carry the same angle $A$. The
same holds for $\cos A$.*
No two different acute angles ever share either ratio. This turns a numerical match into a
statement about angles. If we can show two acute angles share one ratio value, that already shows
the angles themselves are equal.
Your turn: two acute angles have the same $\cos$ value. Are the angles equal? (Answer: yes, since
$\cos$ takes each value at exactly one acute angle.)
Check yourself
As acute angle $A$ runs from $0^\circ$ to $90^\circ$, $\sin A$
falls steadily from $1$ to $0$
rises steadily from $0$ to $1$
rises, then falls back to $0$
stays constant at $1/2$
Check your answer
falls steadily from $1$ to $0$ — Falling from $1$ to $0$ describes $\cos A$; $\sin A$ rises from $0$ to $1$ over the same range.
✓ rises steadily from $0$ to $1$ — (B) $\sin A$ rises steadily from $0$ at $A=0^\circ$ to $1$ at $A=90^\circ$.
rises, then falls back to $0$ — On $0^\circ$ to $90^\circ$, $\sin A$ never turns back down — it rises the whole way to $1$.
stays constant at $1/2$ — $1/2$ is only the value at $A=30^\circ$; across the full range, $\sin A$ keeps rising rather than staying fixed.
Two right triangles give the same value of $\cos A$. What can be concluded about the two angles?
the two angles could be different, since many angles can share the same $\cos A$
the two triangles must be congruent
nothing can be concluded without also comparing $\sin A$
the two angles must be equal, since $\cos A$ never repeats
Check your answer
the two angles could be different, since many angles can share the same $\cos A$ — Because $\cos A$ falls steadily from $1$ to $0$ with no turning point, no two different acute angles share the same value — the angles must be equal.
the two triangles must be congruent — Matching $\cos A$ forces matching angles, not matching triangle sizes — the triangles could still be different sizes, only similar.
nothing can be concluded without also comparing $\sin A$ — $\cos A$ alone is already enough — its steady fall across the range means one value corresponds to exactly one angle.
✓ the two angles must be equal, since $\cos A$ never repeats — (D) Because $\cos A$ falls steadily with no turning point, one value can only ever come from one angle — so the two angles are equal.
A physics problem needs to check whether two measured angles are equal, using only recorded values of $\sin$ for each. Can this check work?
no, because $\sin$ values repeat for several different angles
no, because $\sin$ alone cannot be measured without also knowing $\cos$
yes, since $\sin A$ never repeats a value on the acute range
only if the angles are both less than $45^\circ$
Check your answer
no, because $\sin$ values repeat for several different angles — $\sin A$ never repeats a value on the acute range — it rises steadily from $0$ to $1$, so equal $\sin$ values do mean equal angles.
no, because $\sin$ alone cannot be measured without also knowing $\cos$ — $\sin A$ can be measured on its own, from opposite over hypotenuse — no separate $\cos A$ measurement is needed first.
✓ yes, since $\sin A$ never repeats a value on the acute range — (C) Because $\sin A$ never repeats a value on the acute range, equal recorded $\sin$ values do mean equal angles.
only if the angles are both less than $45^\circ$ — The check works across the whole acute range, $0^\circ$ to $90^\circ$, not just below $45^\circ$.
In the numeric panel, A is the angle opposite the leg of 20, which is why sin A equals 20/29, not 21/29.
One identity ties $\sin A$ and $\cos A$ together, for every acute angle. It follows directly
from Pythagoras’ theorem.
*THEOREM. For any acute angle $A$ of a right triangle, $\sin^2 A + \cos^2 A = 1$.*
We reproduce this one proof in full. Every other identity we meet is built from it.
The numbered proof below divides Pythagoras’ theorem by the hypotenuse squared, then simplifies.
Read it slowly the first time through.
Your turn: for a $5$-$12$-$13$ triangle, does $\sin^2 A + \cos^2 A$ come out to $1$? (Answer: yes.
$(5/13)^2 + (12/13)^2 = 25/169 + 144/169 = 1$.)
Proof
Given. We take a right triangle with acute angle $A$. Call the side opposite $A$ as $p$, the side adjacent to $A$ as $b$, and the hypotenuse as $h$.
To prove. For any acute angle $A$ of a right triangle, $\sin^2 A + \cos^2 A = 1$.
let a right triangle have acute angle $A$, opposite side $p$, adjacent side $b$, hypotenuse $h$ We name the three sides relative to $A$, exactly as in the ratio definitions.
$p^2 + b^2 = h^2$ This is Pythagoras’ theorem, stated earlier.
$p^2/h^2 + b^2/h^2 = h^2/h^2$ We divide both sides of the equation by the same nonzero number, $h^2$. This preserves the equation.
$(p/h)^2 + (b/h)^2 = 1$ This follows from the laws of exponents, since $h^2/h^2 = 1$.
$\sin^2 A + \cos^2 A = 1$ By definition, $p/h = \sin A$ and $b/h = \cos A$.
the identity holds for every acute angle $A$ was any acute angle of any right triangle. Nothing in the argument depended on a particular triangle.
■
Check yourself
The identity $\sin^2 A + \cos^2 A = 1$ holds for
only the angle $A = 45^\circ$
only right triangles with integer side lengths
every acute angle $A$ of a right triangle
only when the triangle is isosceles
Check your answer
only the angle $A = 45^\circ$ — The proof uses an arbitrary right triangle with an arbitrary acute angle $A$ — the identity holds for every one of them, not just $45^\circ$.
only right triangles with integer side lengths — The proof divides by $h^2$ algebraically — nothing in it requires the sides to be integers.
✓ every acute angle $A$ of a right triangle — (C) The proof uses an arbitrary right triangle with an arbitrary acute angle, so the identity holds for every one of them.
only when the triangle is isosceles — The triangle used in the proof is any right triangle with acute angle $A$ — isosceles is never assumed.
The proof of $\sin^2 A + \cos^2 A = 1$ starts from $p^2 + b^2 = h^2$ and divides every term by $h^2$. What does this division achieve?
it finds the numeric value of the hypotenuse $h$
it proves that $p$ and $b$ are equal
it turns the sides into $\sin A$ and $\cos A$
it changes the right angle into an acute angle
Check your answer
it finds the numeric value of the hypotenuse $h$ — The division rewrites the equation in terms of ratios — it does not compute a numeric value for $h$ at all.
it proves that $p$ and $b$ are equal — Nothing about $p$ and $b$ being equal follows from this step — the division produces the ratios $\sin A$ and $\cos A$, whatever $p$ and $b$ happen to be.
✓ it turns the sides into $\sin A$ and $\cos A$ — (C) Dividing by $h^2$ turns each side-length term into a ratio, $(p/h)^2=\sin^2 A$ and $(b/h)^2=\cos^2 A$.
it changes the right angle into an acute angle — The triangle’s angles are untouched by this algebra — dividing by $h^2$ only rewrites the side relationship in terms of ratios.
Why does the proof of $\sin^2 A + \cos^2 A = 1$ use Pythagoras’ theorem rather than measuring $\sin A$ and $\cos A$ directly in many triangles and checking the sum?
because measuring $\sin A$ and $\cos A$ directly is impossible in any triangle
because Pythagoras’ theorem gives a more accurate numeric answer than measurement
because measurement only works for angles below $45^\circ$
because measurement only ever checks finitely many triangles
Check your answer
because measuring $\sin A$ and $\cos A$ directly is impossible in any triangle — $\sin A$ and $\cos A$ can certainly be measured in any one triangle — the problem is that measuring finitely many triangles never proves the identity for every acute angle.
because Pythagoras’ theorem gives a more accurate numeric answer than measurement — The point is not measurement accuracy — it is that an algebraic proof covers every acute angle at once, while any set of measurements covers only the triangles actually measured.
because measurement only works for angles below $45^\circ$ — Measurement is not restricted to angles below $45^\circ$ — the real limitation is that it can never cover every angle at once, which is exactly what the algebraic proof does.
✓ because measurement only ever checks finitely many triangles — (D) An algebraic proof covers every acute angle at once; measuring even many triangles only ever checks the finitely many that were measured.
In a right triangle, $\sin A = 0.6$. Using $\sin^2 A + \cos^2 A = 1$, what is $\cos A$ (taking the positive root)?
The same Pythagoras relation, $p^2 + b^2 = h^2$, gives two more identities by dividing
by a different squared side.
Divide by $b^2$, the adjacent side’s square, and we reach $1 + \tan^2 A = \sec^2 A$. Divide the same
relation by $p^2$, the opposite side’s square. The result is $1 + \cot^2 A = \text{cosec}^2 A$
instead.
$1 + \tan^2 A = \sec^2 A$ holds for every acute $A$. $1 + \cot^2 A = \text{cosec}^2 A$ needs $\cot A$ and
$\text{cosec} A$ to be defined. This rules out $A = 0^\circ$.
Your turn: dividing $p^2 + b^2 = h^2$ by $p^2$ gives which identity? (Answer:
$1 + \cot^2 A = \text{cosec}^2 A$.)
All three identities are one theorem divided three ways: Pythagoras’ relation shared out by the square of whichever side you choose.
Check yourself
Dividing $p^2+b^2=h^2$ by $b^2$, the square of the adjacent side, gives
$1 + \cot^2 A = \text{cosec}^2 A$
$\sin^2 A + \cos^2 A = 1$
$1 - \tan^2 A = \sec^2 A$
$1 + \tan^2 A = \sec^2 A$
Check your answer
$1 + \cot^2 A = \text{cosec}^2 A$ — Dividing by $p^2$ instead gives $1+\cot^2 A = \text{cosec}^2 A$; dividing by $b^2$ gives $1+\tan^2 A = \sec^2 A$.
$\sin^2 A + \cos^2 A = 1$ — That is the identity before this division; dividing by $b^2$ produces a new one, $1+\tan^2 A = \sec^2 A$.
$1 - \tan^2 A = \sec^2 A$ — Every term in $p^2+b^2=h^2$ stays added after dividing by $b^2$ — the identity is $1+\tan^2 A = \sec^2 A$, with a plus.
✓ $1 + \tan^2 A = \sec^2 A$ — (D) Dividing $p^2+b^2=h^2$ by $b^2$ gives $(p/b)^2+1=(h/b)^2$, which is $\tan^2 A+1=\sec^2 A$.
Both $1+\tan^2 A = \sec^2 A$ and $1+\cot^2 A = \text{cosec}^2 A$ come from the same source. What is it?
two separate, unrelated proofs, one for each identity
measurement of many right triangles
the standard-angle table’s exact values
Pythagoras’ theorem, divided differently each time
Check your answer
two separate, unrelated proofs, one for each identity — Both identities come from the same equation, $p^2+b^2=h^2$, just divided by a different side’s square — they are not separately proved.
measurement of many right triangles — Both identities are derived algebraically from Pythagoras’ theorem, not measured from triangles.
the standard-angle table’s exact values — The standard-angle table lists specific numbers at five angles; these identities hold for every acute angle, derived algebraically.
✓ Pythagoras’ theorem, divided differently each time — (D) Both identities come from Pythagoras’ theorem, $p^2+b^2=h^2$, just divided by a different side’s square each time.
Given $\tan A = 5/12$, use $1+\tan^2 A = \sec^2 A$ to find $\sec A$ (taking the positive root).
$17/12$
$12/13$
$169/144$
$13/12$
Check your answer
$17/12$ — The identity needs $\tan^2 A$, not $\tan A$ itself: $1+(5/12)^2 = 1+25/144 = 169/144$, whose root is $13/12$.
$12/13$ — $12/13$ is $\cos A$, the reciprocal of $\sec A$; $\sec A$ itself is $13/12$.
$169/144$ — $169/144$ is $\sec^2 A$, not $\sec A$ itself — the square root still needs taking, giving $13/12$.
Given cos A equals 5/13, the identity forces sin A equals 12/13, taking the positive root for an acute angle.
Once we know one ratio of an acute angle, the Pythagorean identity gives every other ratio
algebraically. No triangle needs to be drawn at all.
From $\sin A$ alone, $\cos A = \sqrt{1 - \sin^2 A}$. Take the positive root: you already know every
ratio of an acute angle is positive. Every other ratio then follows from these two, using the reciprocal and
quotient relations already met.
An acute angle never gives a negative ratio. The negative root is never the answer here.
Your turn: given $\cos A = 5/13$, what is $\sin A$? (Answer: $\sin A = \sqrt{1 - 25/169} = 12/13$.)
Check yourself
Given only $\sin A$ for an acute angle $A$, every other ratio can be found because
$\cos A = \sqrt{1-\sin^2 A}$ gives every other ratio
a triangle must first be drawn and measured with a ruler
$\tan A$ must be looked up separately in the standard-angle table
$\cos A$ and $\sin A$ are actually independent and unrelated
Check your answer
✓ $\cos A = \sqrt{1-\sin^2 A}$ gives every other ratio — (A) $\cos A = \sqrt{1-\sin^2 A}$ follows from the identity, and every other ratio is then built from $\sin A$ and $\cos A$.
a triangle must first be drawn and measured with a ruler — No drawing or measurement is needed — the identity gives $\cos A$ algebraically, and every other ratio follows from $\sin A$ and $\cos A$.
$\tan A$ must be looked up separately in the standard-angle table — This method works for any acute angle, not only the five in the standard table — $\tan A$ is computed as $(\sin A)/(\cos A)$, never looked up.
$\cos A$ and $\sin A$ are actually independent and unrelated — $\sin A$ and $\cos A$ are linked by $\sin^2 A + \cos^2 A = 1$ — they are not independent, which is exactly why one determines the other.
Why is only the positive square root taken when writing $\cos A = \sqrt{1-\sin^2 A}$ for an acute angle $A$?
because square roots are always defined as positive by convention, regardless of context
because $1-\sin^2 A$ is always a perfect square
because $A$ is acute, so $\cos A$ must be positive, and the negative root would be wrong
because $\cos A$ is always larger than $\sin A$
Check your answer
because square roots are always defined as positive by convention, regardless of context — Square roots can carry a negative sign in general — the positive root is chosen here specifically because $A$ is acute, so $\cos A$ must be positive.
because $1-\sin^2 A$ is always a perfect square — $1-\sin^2 A$ need not be a perfect square at all — the root is taken positive because of the angle being acute, not because of any special numeric property.
✓ because $A$ is acute, so $\cos A$ must be positive, and the negative root would be wrong — (C) $A$ being acute forces $\cos A$ to be positive, so only the positive root is kept.
because $\cos A$ is always larger than $\sin A$ — $\cos A$ is not always larger than $\sin A$ — past $45^\circ$, $\sin A$ overtakes $\cos A$. The positive root is chosen because $A$ is acute, not because of relative size.
Given $\sin A = 5/13$ for an acute angle $A$, find $\tan A$ using $\cos A = \sqrt{1-\sin^2 A}$.
$5/13$
$12/13$
$5/12$
$12/5$
Check your answer
$5/13$ — $5/13$ is the given $\sin A$; $\tan A$ still needs computing as $(\sin A)/(\cos A) = 5/12$.
$12/13$ — $12/13$ is $\cos A = \sqrt{1-(5/13)^2}$; $\tan A$ is $\sin A$ divided by that, $5/12$.
✓ $5/12$ — (C) $\cos A = \sqrt{1-(5/13)^2}=12/13$, so $\tan A = (5/13)/(12/13)=5/12$.
$12/5$ — $12/5$ is $\cot A$, the reciprocal of $\tan A$; $\tan A$ itself is $5/12$.
There is a tempting shortcut when proving an identity. Manipulate both sides at once, as
though they were already equal. Cross-multiply. Simplify. Stop once the two sides match.
That method proves nothing. If we treat an unproved identity as already true, we assume the very
thing the proof was meant to show. The match at the end was fixed from the first step, not earned.
A genuine proof works from one side alone, using only identities already established, until that
side becomes the other.
The statement being proved never appears anywhere in the middle of the working. The worked example
proving $\sec A (1 - \sin A)(\sec A + \tan A) = 1$ follows exactly this rule.
Your turn: a working assumes $\sec A (1 - \sin A)(\sec A + \tan A) = 1$ from the start. Then
it simplifies both sides until they match. Is this a valid proof? (Answer: no. It assumes the
very thing it set out to prove.)
First try
To prove $1 + \tan^2 A = \sec^2 A$ for $A = 30^\circ$, I assumed the two sides equal, then simplified both sides together until they matched. $\sin^2 A + \cos^2 A = 1$ came out at the end, so I called the identity proved.
Second look
That assumes the very thing it sets out to prove. Working from the left side alone: $1 + \tan^2 A = 1 + (\sin^2 A)/(\cos^2 A) = (\cos^2 A + \sin^2 A)/(\cos^2 A) = 1/(\cos^2 A) = \sec^2 A$. For $A = 30^\circ$, this gives $1 + 1/3 = 4/3$, exactly $\sec^2 30^\circ$.
Work from one side alone. The identity being proved must never appear until the very last line.
Does working on both sides at once ever prove an identity?
Weaker. We want to prove $1 + \tan^2 A = \sec^2 A$. Start by assuming both sides are equal. Write
$1 + (\sin A)^2/(\cos A)^2 = 1/(\cos A)^2$. Subtract $1$ from both sides:
$(\sin A)^2/(\cos A)^2 = 1/(\cos A)^2 - 1$. Multiply both sides by $\cos^2 A$: $\sin^2 A = 1 - \cos^2 A$.
This is just $\sin^2 A + \cos^2 A = 1$, already known. The two sides match, so the identity is
“proved.”
The catch: line two already treated $1 + \tan^2 A = \sec^2 A$ as true, before it was shown.
Everything after that just restates a fact already known. Nothing here shows the identity itself
is true.
Stronger. We want to prove $1 + \tan^2 A = \sec^2 A$, working from the left side alone.
$1 + \tan^2 A = 1 + (\sin^2 A)/(\cos^2 A)$, writing $\tan A$ in terms of $\sin A$ and $\cos A$. Combine
over one denominator: $= (\cos^2 A + \sin^2 A)/(\cos^2 A)$. Use $\sin^2 A + \cos^2 A = 1$ once, in the
numerator: $= 1/(\cos^2 A)$. This is $\sec^2 A$, the right side.
The identity being proved never appeared until the very last line, where the left side finished
equal to the right side.
A SCHOLAR INDIA REMEMBERS
You wrote the identity at the top, then showed it was true. Did you earn the answer, or did you assume it? Start from one side only. Change it one step at a time. Stop when it has become the other side. Now the identity is earned.
Aryabhata · the hoopoe
Check yourself
A student proves $\sec A - \tan A \cdot \sin A = \cos A$ by writing: “Cross-multiplying both sides, $\sec A \cdot \cos A - \tan A \cdot \sin A \cdot \cos A = \cos A \cdot \cos A$, which simplifies to $1 - \sin^2 A = \cos^2 A$, true by the Pythagorean identity, so the original statement is proved.” What is wrong with this method?
nothing is wrong — manipulating both sides of an identity at once is a standard, valid way to confirm it
the final simplified equation, $1-\sin^2 A = \cos^2 A$, is actually false
$\sec A$ was defined incorrectly in the working
it manipulates both sides of the identity at once, as if they were already known to be equal
Check your answer
nothing is wrong — manipulating both sides of an identity at once is a standard, valid way to confirm it — Manipulating both sides at once assumes they are already equal — exactly the thing an identity proof must establish, not assume; it is not a valid technique here.
the final simplified equation, $1-\sin^2 A = \cos^2 A$, is actually false — $1-\sin^2 A = \cos^2 A$ is a true identity on its own — the flaw is that the student worked on both sides together, assuming the very equality being proved.
$\sec A$ was defined incorrectly in the working — $\sec A$ is used correctly throughout — the error is structural: the proof assumes both sides are equal from the start, instead of working from one side alone.
✓ it manipulates both sides of the identity at once, as if they were already known to be equal — (D) The method treats the two sides as already equal by working on both at once — exactly the equality the proof is meant to establish.
Why does “manipulate both sides at once until they match” fail as a proof method for an identity, even when every algebraic step is individually correct?
because algebra performed on both sides of an equation is never actually valid
because it only works for identities involving $\sin$ and $\cos$, not other ratios
because it assumes the very equality it sets out to prove
because it takes too many steps compared to working from one side alone
Check your answer
because algebra performed on both sides of an equation is never actually valid — Operating on both sides of a known equation is standard and valid — the problem here is treating an unproven identity as already known, before it has been established.
because it only works for identities involving $\sin$ and $\cos$, not other ratios — The flaw applies to any identity proved this way, regardless of which ratios appear in it — the issue is the method, not the specific ratios.
✓ because it assumes the very equality it sets out to prove — (C) Even with every step individually correct, the method assumes the equality it is meant to prove, which is the flaw.
because it takes too many steps compared to working from one side alone — The problem is not the number of steps — a proof that assumes what it sets out to show is invalid however short it is.
To prove $(1-\cos A)/(\sin A) = (\sin A)/(1+\cos A)$, a valid method works from
both sides at once, cross-multiplying and simplifying until the two sides visibly match
substituting specific numeric values of $A$, such as $A=30^\circ$, to check the identity holds
assuming the identity is true and using it to simplify a different, unrelated expression
one side alone, rewritten using identities already known
Check your answer
both sides at once, cross-multiplying and simplifying until the two sides visibly match — Cross-multiplying both sides and simplifying until they match is exactly the circular method that assumes the identity to prove it — a valid proof starts from one side alone.
substituting specific numeric values of $A$, such as $A=30^\circ$, to check the identity holds — Checking one angle, such as $30^\circ$, only confirms that single case — a proof must work algebraically for every acute angle, not one number.
assuming the identity is true and using it to simplify a different, unrelated expression — Assuming the identity true anywhere in the proof, even to simplify something else, repeats the same circularity the method must avoid.
✓ one side alone, rewritten using identities already known — (D) A valid proof rewrites one side alone, using established identities, until it becomes the other side.
A worked example in this chapter proves $\sec A(1-\sin A)(\sec A+\tan A)=1$ by rewriting only the left side in terms of $\sin A$ and $\cos A$ until it becomes $1$. Why does this method count as valid, unlike manipulating both sides at once?
because the left side happens to be more complicated than the right side
because $\sec A$ and $\tan A$ are reciprocal ratios, which makes any proof about them automatically valid
because the final answer, $1$, is a particularly simple number to reach
because it never assumes the equality it is proving
Check your answer
because the left side happens to be more complicated than the right side — Which side looks more complicated is irrelevant — what makes the method valid is that it never assumes the equality it is proving, working from one side alone.
because $\sec A$ and $\tan A$ are reciprocal ratios, which makes any proof about them automatically valid — Being a reciprocal ratio does not make a proof automatically valid — validity comes from the method, working from one side alone, not from which ratios are involved.
because the final answer, $1$, is a particularly simple number to reach — Reaching a simple final number does not make a method valid — what matters is that the working never assumed the equality it set out to prove.
✓ because it never assumes the equality it is proving — (D) The method never assumes the equality it is proving — it only rewrites one side, using identities already known to hold.
$\sin A$ looks, at a glance, like two things side by side. It is not “sin” multiplied by
$A$. Read that way, a step like $\sin A \cdot \cos A$ might tempt us to cancel the $A$’s. That attempt
means nothing.
$\sin A$, $\cos A$, $\tan A$, $\text{cosec} A$, $\sec A$ and $\cot A$ are each one inseparable symbol. Each is read as “the sine of
angle $A$,” and so on. “Sin” written alone, with no angle attached, names nothing at all. It can never be split
off, multiplied, cancelled, or factored on its own.
Your turn: in $(\sin A)/(\cos A)$, is it valid to cancel the two $A$’s? (Answer: no. $\sin A$ and
$\cos A$ are each one symbol; there is no lone $A$ to cancel.)
In sine of A divided by sine of B, can the sine be cancelled, leaving A over B?
Weaker. A learner sees $(\sin A)/(\sin B)$ and reads it as “sin” times $A$, divided by “sin” times $B$.
Cancelling the “sin” from top and bottom, they write $(\sin A)/(\sin B) = A/B$. For $A = 30^\circ$
and $B = 60^\circ$, this gives $A/B = 1/2$.
Check it against real values: $\sin 30^\circ = 1/2$ and $\sin 60^\circ = \sqrt{3}/2$, so
$(\sin A)/(\sin B) = (1/2)/(\sqrt{3}/2) = 1/\sqrt{3}$, not $1/2$. The cancelled “sin” was never a
separate factor to cancel: there is no lone “sin” standing anywhere in $(\sin A)/(\sin B)$.
Stronger. $\sin A$ and $\sin B$ are each one number, not a product. $(\sin A)/(\sin B)$ is one number divided
by another, and nothing in it cancels. For $A = 30^\circ$ and $B = 60^\circ$,
$(\sin A)/(\sin B) = (1/2)/(\sqrt{3}/2) = 1/\sqrt{3}$.
That is the actual value, found by computing each sine first, then dividing. No letter or symbol
was cancelled at any step.
Check yourself
A student simplifies $\sin A / A$ by “cancelling” the $A$, writing the result as $\sin$. What is wrong with this step?
the step is valid, since $\sin A$ really does mean “sin” multiplied by $A$
the student used the wrong ratio; the step would be valid for $\tan A / A$ instead
the student forgot to convert $A$ from degrees to radians before cancelling
$\sin A$ is one inseparable symbol, not “sin” times $A$
Check your answer
the step is valid, since $\sin A$ really does mean “sin” multiplied by $A$ — $\sin A$ is never “sin” multiplied by $A$ — it is one inseparable symbol for “the sine of angle $A$”, so no such cancellation is ever valid.
the student used the wrong ratio; the step would be valid for $\tan A / A$ instead — The same error would occur with $\tan A / A$ too — no ratio’s name can be split from its angle and cancelled, whichever ratio is involved.
the student forgot to convert $A$ from degrees to radians before cancelling — Converting units would not fix this — the real issue is that “sin” alone, split from $A$, has no meaning to cancel with anything.
✓ $\sin A$ is one inseparable symbol, not “sin” times $A$ — (D) $\sin A$ is one inseparable symbol; “sin” alone has no meaning that could be cancelled or multiplied.
Which of these is a correct way to read the symbol $\sin A$?
“sin” multiplied by $A$
“the sine of angle $A$”, one inseparable symbol
the letter “sin” raised to the power $A$
the value of $A$ measured in a unit called “sin”
Check your answer
“sin” multiplied by $A$ — “sin” alone names nothing that could be multiplied by $A$ — $\sin A$ is one symbol, not a product of two.
✓ “the sine of angle $A$”, one inseparable symbol — (B) $\sin A$ reads as “the sine of angle $A$” — one inseparable symbol, never two pieces multiplied together.
the letter “sin” raised to the power $A$ — There is no exponent here — $\sin A$ is a single named ratio, not “sin” raised to a power.
the value of $A$ measured in a unit called “sin” — “Sin” is not a unit of measurement — $\sin A$ names a specific ratio of two sides, tied to the angle $A$.
Why can “sin” never be split off from the angle $A$ and treated as a separate factor?
because “sin” is a reserved word in mathematics that cannot legally appear alone in an expression
because splitting it would make the expression too long to simplify
because “sin” alone names no number or quantity
because $A$ must always be written in degrees, never split from its unit
Check your answer
because “sin” is a reserved word in mathematics that cannot legally appear alone in an expression — This is not a rule against writing “sin” alone — it is that “sin” alone simply names no number or quantity to work with.
because splitting it would make the expression too long to simplify — Length is not the issue — “sin” alone has no numeric value to separate out, regardless of how long or short the surrounding expression is.
✓ because “sin” alone names no number or quantity — (C) “Sin” alone names nothing — it only becomes a value once paired with a specific angle.
because $A$ must always be written in degrees, never split from its unit — This has nothing to do with degrees versus radians — “sin” cannot be split off from $A$ because “sin” alone is meaningless, in any unit.
A student is asked to evaluate $2 \sin A$ for $A = 30^\circ$ and writes the answer as $2 \cdot \sin \cdot 30^\circ = 60^\circ \cdot \sin$. What went wrong?
$\sin A$ was wrongly split as a multiplication
the student forgot to convert $30^\circ$ into radians first
the student should have used $\cos$ instead of $\sin$ for this angle
the final answer of $1$ is wrong; it should be a larger number
Check your answer
✓ $\sin A$ was wrongly split as a multiplication — (A) $\sin A$ was wrongly split apart as a multiplication; read correctly, $2 \sin 30^\circ = 2 \cdot (1/2) = 1$.
the student forgot to convert $30^\circ$ into radians first — No radian conversion is required here — the mistake is treating $\sin A$ as a product; $2 \sin 30^\circ = 2 \cdot (1/2) = 1$.
the student should have used $\cos$ instead of $\sin$ for this angle — $\sin$ is the correct ratio asked for — the error is not which ratio was chosen, but treating $\sin A$ as a multiplication.
the final answer of $1$ is wrong; it should be a larger number — $1$ is exactly the correct value of $2 \sin 30^\circ$ — the actual mistake is the multiplication step, not the size of the final number.
$\tan A = 4/3$, so opposite $= 4 k$ and adjacent $= 3 k$ for some positive $k$ $\tan A$ is opposite over adjacent. The given ratio fixes these two sides, up to a scale factor $k$.
hypotenuse $= \sqrt{(4 k)^2 + (3 k)^2} = \sqrt{25 k^2} = 5 k$ Pythagoras’ theorem supplies the third side.
$\sin A = 4/5$, $\cos A = 3/5$, $\tan A = 4/3$ We read the three ratios directly off the three sides. The $k$ cancels in every ratio.
$\text{cosec} A = 5/4$, $\sec A = 5/3$, $\cot A = 3/4$ Each is the reciprocal of the matching ratio above.
$(4/5)^2 + (3/5)^2 = 16/25 + 9/25 = 1$ *Check by squaring $\sin A$ and $\cos A$ and adding them.* They must add to $1$, and here they do.
Keeping the scale factor attached to every side is why these three ratios stay identical no matter how large the triangle is built.
Given the tangent of an angle, find the other five ratios.
Scale into two sides For $\tan A = p/q$, set opposite $= p k$ and adjacent $= q k$, for some positive $k$. Here $\tan A = 4/3$ gives opposite $= 4 k$, adjacent $= 3 k$.
Find the hypotenuse Use Pythagoras’ theorem, hypotenuse $= \sqrt{(p k)^2 + (q k)^2}$. Here that is $\sqrt{(4 k)^2 + (3 k)^2} = 5 k$.
Read off sine and cosine $\sin A$ is opposite over hypotenuse; $\cos A$ is adjacent over hypotenuse. Here $\sin A = 4/5$ and $\cos A = 3/5$.
Take the three reciprocals $\text{cosec} A = 1/(\sin A)$, $\sec A = 1/(\cos A)$, $\cot A = 1/(\tan A)$. Here that gives $5/4$, $5/3$ and $3/4$.
Check against the identity Square $\sin A$ and $\cos A$ and add them. They must equal $1$. Here $(4/5)^2 + (3/5)^2 = 1$.
Check yourself
Given $\tan A = 4/3$ for an acute angle $A$, the standard recipe writes the opposite and adjacent sides as
$4$ and $3$, exactly, with no scaling factor
$4k$ and $3k$, for some positive $k$
$3k$ and $4k$, with adjacent listed first
$4k$ and $3k$, naming the hypotenuse and one leg
Check your answer
$4$ and $3$, exactly, with no scaling factor — The sides could be any multiple of $4$ and $3$, such as $8$ and $6$ — the scaling factor $k$ keeps this general.
✓ $4k$ and $3k$, for some positive $k$ — (B) Opposite $=4k$ and adjacent $=3k$, for some positive $k$, matching the numerator and denominator of $\tan A$.
$3k$ and $4k$, with adjacent listed first — $\tan A$ is opposite over adjacent, so the numerator $4$ names the opposite side and the denominator $3$ the adjacent side — opposite is $4k$, adjacent is $3k$.
$4k$ and $3k$, naming the hypotenuse and one leg — $\tan A$ never involves the hypotenuse at all — both $4k$ and $3k$ are legs; the hypotenuse is found afterward using Pythagoras’ theorem.
Given opposite $=4k$ and adjacent $=3k$, the hypotenuse is found as $\sqrt{(4k)^2+(3k)^2} = 5k$. Why is $5k$ specifically, and not another multiple of $k$, the correct hypotenuse?
because every right triangle with legs in ratio $4$ to $3$ must have hypotenuse $5$ units long
because $(4k)^2+(3k)^2 = 25k^2$, and its square root is exactly $5k$
because $4+3=7$, and $5$ is roughly half of $7$
because $5$ is the next prime number after $4$ and $3$
Check your answer
because every right triangle with legs in ratio $4$ to $3$ must have hypotenuse $5$ units long — The hypotenuse scales with $k$ just like the legs — it is $5k$, not a fixed $5$ units, whatever value $k$ takes.
✓ because $(4k)^2+(3k)^2 = 25k^2$, and its square root is exactly $5k$ — (B) $(4k)^2+(3k)^2=16k^2+9k^2=25k^2$, and $\sqrt{25k^2}=5k$.
because $4+3=7$, and $5$ is roughly half of $7$ — The hypotenuse comes from $\sqrt{(4k)^2+(3k)^2}$, not from adding and halving the two leg numbers — that shortcut has no mathematical basis.
because $5$ is the next prime number after $4$ and $3$ — Primality plays no role here — $5k$ comes directly from squaring, adding and rooting the two legs, $\sqrt{16k^2+9k^2}=5k$.
Given $\tan A = 4/3$, the recipe gives opposite $=4k$, adjacent $=3k$, hypotenuse $=5k$. What is $\sec A$?
$3/5$
$5/3$
$5/4$
$4/3$
Check your answer
$3/5$ — $3/5$ is $\cos A$; $\sec A$ is its reciprocal, $5/3$.
✓ $5/3$ — (B) $\sec A =$ hypotenuse over adjacent $= 5k/3k = 5/3$.
$5/4$ — $5/4$ is $\text{cosec} A$ (hypotenuse over opposite); $\sec A$ uses the adjacent side, $5/3$.
$4/3$ — $4/3$ is the given $\tan A$; $\sec A$ is a different ratio, hypotenuse over adjacent, $5/3$.
Checking the identity on whole numbers checks that two parts fill a whole: 400 and 441 make 841, so the squared ratios make exactly 1.
Worked example
Checking the identity on a 20-21-29 triangle
hypotenuse $29$, one leg $21$, so the other leg is $\sqrt{29^2 - 21^2} = \sqrt{841 - 441} = \sqrt{400} = 20$ Pythagoras’ theorem, rearranged to find the missing leg.
for the angle $A$ opposite the leg $20$: $\sin A = 20/29$, $\cos A = 21/29$ We read the ratios off the three sides of this triangle.
$(20/29)^2 + (21/29)^2 = 400/841 + 441/841 = 841/841 = 1$ *This confirms $\sin^2 A + \cos^2 A = 1$ numerically*, on a triangle where every side is a whole number.
Check the Pythagorean identity on a right triangle with whole-number sides.
Confirm the triangle is right-angled The identity only holds for a right triangle. Check a right angle is marked before using Pythagoras’ theorem.
Find the side not given Use $c^2 = a^2 + b^2$ to find the missing side. For hypotenuse $29$ and leg $21$, the missing leg is $\sqrt{29^2 - 21^2} = 20$.
Read sine and cosine off the three sides Pick the angle $A$ at one acute vertex. $\sin A$ is opposite over hypotenuse; $\cos A$ is adjacent over hypotenuse. Here $\sin A = 20/29$, $\cos A = 21/29$.
Square and add Check $\sin^2 A + \cos^2 A = 1$ using the fractions from the last step. Here $(20/29)^2 + (21/29)^2 = 1$.
Check yourself
A right triangle has hypotenuse $29$ and one leg $21$. Using Pythagoras’ theorem, the other leg is
$\sqrt{50}$
$8$
$20$
$50$
Check your answer
$\sqrt{50}$ — The missing leg comes from $\sqrt{29^2-21^2}=\sqrt{400}=20$; the hypotenuse’s square must be reduced by the known leg’s square, not added to it.
$8$ — $29-21=8$ skips the theorem entirely — the correct method squares each side first: $\sqrt{29^2-21^2}=20$.
$50$ — $29+21=50$ is not how the missing leg is found — Pythagoras’ theorem needs $\sqrt{29^2-21^2}=20$.
For the triangle with sides $20$, $21$ and hypotenuse $29$, $\sin A = 20/29$ and $\cos A = 21/29$ give $(20/29)^2+(21/29)^2=1$. What does this numeric check confirm?
it confirms the identity in this one triangle
that the identity is proved true for every acute angle by this one example
that $20$, $21$ and $29$ are the only side lengths for which the identity holds
that $\sin A$ and $\cos A$ must always add to $1$, not just their squares
Check your answer
✓ it confirms the identity in this one triangle — (A) This one numeric case is consistent with the general proof — it confirms the identity here, without replacing the general argument.
that the identity is proved true for every acute angle by this one example — One numeric example, however clean, only checks one triangle — the general proof (dividing $p^2+b^2=h^2$ by $h^2$) is what covers every acute angle.
that $20$, $21$ and $29$ are the only side lengths for which the identity holds — The identity holds for every right triangle’s sides, not only $20$-$21$-$29$ — this triple is just one convenient numeric check.
that $\sin A$ and $\cos A$ must always add to $1$, not just their squares — It is the squares that sum to $1$; $\sin A + \cos A$ itself does not generally equal $1$ — here it is $41/29$, not $1$.
For the same $20$-$21$-$29$ triangle, what is $\tan A$ for the angle opposite the side of length $20$?
$21/20$
$20/21$
$20/29$
$29/21$
Check your answer
$21/20$ — $21/20$ is $\cot A$, the reciprocal of $\tan A$; $\tan A$ itself is $20/21$.
✓ $20/21$ — (B) $\tan A =$ opposite over adjacent $= 20/21$.
$20/29$ — $20/29$ is $\sin A$; $\tan A$ is opposite over adjacent, $20/21$, not opposite over hypotenuse.
$29/21$ — $29/21$ is $\sec A$ (hypotenuse over adjacent); $\tan A$ is opposite over adjacent, $20/21$.
in $\text{ABC}$, right-angled at $B$, angle $C = 30^\circ$ and $\text{AB} = 5$ cm This is the given information. The right angle and the known angle are both named.
$\text{AB}$ is opposite $C$ and $\text{BC}$ is adjacent to $C$ We name the sides relative to the angle we are given, $C$, not to the right angle at $B$.
$\tan 30^\circ = \text{AB}/\text{BC}$, so $\text{BC} = \text{AB}/(\tan 30^\circ) = 5 \sqrt{3}$ cm $\tan 30^\circ = 1/\sqrt{3}$, from the standard-angle table.
$\sin 30^\circ = \text{AB}/\text{AC}$, so $\text{AC} = \text{AB}/(\sin 30^\circ) = 10$ cm $\sin 30^\circ = 1/2$, from the same table.
$5^2 + (5 \sqrt{3})^2 = 25 + 75 = 100 = 10^2$ Check both sides found against Pythagoras’ theorem. They should give the hypotenuse back, and here they do.
the ladder
the wall
flat ground
the angle with the wall
The ladder is the hypotenuse. The wall and the flat ground are the triangle’s other two sides.
The same two ratios used here to solve for the missing sides are the ones most missing-side problems reach for first.
Find two unknown sides using the standard-angle table.
Name the sides relative to the given angle Call the side opposite the given angle “opposite,” and the side touching it that is not the hypotenuse “adjacent.” For angle $C = 30^\circ$ with $\text{AB} = 5$ cm opposite $C$, $\text{BC}$ is adjacent to $C$.
Pick a ratio linking a known side to an unknown one Read the ratio for the given angle off the standard-angle table. Here $\tan 30^\circ = \text{AB}/\text{BC}$ gives $\text{BC} = 5 \sqrt{3}$ cm.
Pick a second ratio for the last side Use a different ratio at the same angle. Here $\sin 30^\circ = \text{AB}/\text{AC}$ gives $\text{AC} = 10$ cm.
Check with Pythagoras’ theorem The three sides found must satisfy $a^2 + b^2 = c^2$. Here $5^2 + (5 \sqrt{3})^2 = 10^2$.
Check yourself
In triangle $\text{ABC}$, right-angled at $B$, with angle $C=30^\circ$ and $\text{AB}=5$ cm, the side $\text{AB}$ is
adjacent to angle $C$
the hypotenuse of the triangle
opposite to angle $C$
equal in length to $\text{BC}$
Check your answer
adjacent to angle $C$ — $\text{AB}$ faces angle $C$ directly, making it the opposite side; $\text{BC}$, next to angle $C$, is the adjacent side.
the hypotenuse of the triangle — The hypotenuse is opposite the right angle at $B$, which is $\text{AC}$ — $\text{AB}$ is a leg, opposite angle $C$.
✓ opposite to angle $C$ — (C) $\text{AB}$ faces angle $C$ directly, making it the side opposite $C$.
equal in length to $\text{BC}$ — Nothing states $\text{AB}$ and $\text{BC}$ are equal — in fact $\text{BC}=5\sqrt{3}$ cm while $\text{AB}=5$ cm, given.
In the same triangle, $\tan 30^\circ = \text{AB}/\text{BC}$ is used to find $\text{BC}$. Why is $\tan$, rather than $\sin$ or $\cos$, the right ratio to use here?
because $\tan$ always gives a larger, more accurate answer than $\sin$ or $\cos$
because the angle given is exactly $30^\circ$, and $\tan$ only works for that angle
$\tan$ links the known opposite side to the wanted adjacent side
because $\sin$ and $\cos$ cannot be used when the hypotenuse is unknown
Check your answer
because $\tan$ always gives a larger, more accurate answer than $\sin$ or $\cos$ — Accuracy is not the issue — $\tan$ is chosen because it relates opposite and adjacent directly, without the hypotenuse that is not yet known.
because the angle given is exactly $30^\circ$, and $\tan$ only works for that angle — $\tan$ works for any acute angle, not only $30^\circ$ — it is chosen here because it links the two sides in question, opposite and adjacent.
✓ $\tan$ links the known opposite side to the wanted adjacent side — (C) $\tan$ links the given opposite side directly to the wanted adjacent side, without ever needing the (unknown) hypotenuse.
because $\sin$ and $\cos$ cannot be used when the hypotenuse is unknown — $\sin$ and $\cos$ could still be used together with more steps — $\tan$ is simply the most direct choice, since it skips the hypotenuse entirely.
Continuing the same triangle, $\sin 30^\circ = \text{AB}/\text{AC}$ gives $\text{AC}=10$ cm. Using $\text{AB}=5$ cm and $\text{BC}=5\sqrt{3}$ cm, does this value of $\text{AC}$ agree with Pythagoras’ theorem?
no, since $5+5\sqrt{3}$ does not equal $10$
yes, since $5^2+(5\sqrt{3})^2 = 25+75 = 100 = 10^2$
no, since the angle $30^\circ$ was not used in the check
it cannot be checked without knowing angle $A$ as well
Check your answer
no, since $5+5\sqrt{3}$ does not equal $10$ — Pythagoras’ theorem checks squares, not a direct sum: $5^2+(5\sqrt{3})^2=100=10^2$ agrees exactly.
✓ yes, since $5^2+(5\sqrt{3})^2 = 25+75 = 100 = 10^2$ — (B) $5^2+(5\sqrt{3})^2=25+75=100=10^2$, so $\text{AC}=10$ cm agrees exactly.
no, since the angle $30^\circ$ was not used in the check — Pythagoras’ theorem only needs the three side lengths, already found using the angle — the angle does not need to appear again in the check itself.
it cannot be checked without knowing angle $A$ as well — The three side lengths alone are enough for a Pythagoras check — no further angle is needed.
$\tan A = 1$, so the two legs of the right triangle are equal $\tan A$ is opposite over adjacent. That ratio is $1$ only when the two legs match.
let each leg be $k$, so the hypotenuse is $\sqrt{k^2 + k^2} = k \sqrt{2}$ Pythagoras’ theorem, with both legs equal to $k$.
$\sin A = k/(k \sqrt{2}) = 1/\sqrt{2}$, $\cos A = 1/\sqrt{2}$ We read both ratios off the three sides. The $k$ cancels.
$2 \cdot \sin A \cdot \cos A = 2 \cdot (1/\sqrt{2}) \cdot (1/\sqrt{2}) = 2/2 = 1$ *Check by substituting both ratios back into $2 \sin A \cos A$.* It simplifies to exactly $1$.
Check yourself
If $\tan A = 1$ for an acute angle $A$ in a right triangle, the two legs are
in the ratio $2$ to $1$
equal in length
always exactly $1$ unit long
impossible, since no two sides of a right triangle can be equal
Check your answer
in the ratio $2$ to $1$ — $\tan A = 1$ means opposite over adjacent equals $1$, so the two legs must be exactly equal, not in a $2$-to-$1$ ratio.
✓ equal in length — (B) $\tan A=1$ means opposite over adjacent equals $1$, so the two legs are equal.
always exactly $1$ unit long — $\tan A = 1$ fixes the ratio of the legs, not their absolute size — the legs could be $1$ unit each, or $5$ units each, as long as they are equal.
impossible, since no two sides of a right triangle can be equal — A right triangle with two equal legs, a right isosceles triangle, is perfectly possible — it is exactly what $\tan A=1$ describes.
With both legs equal to $k$, the hypotenuse works out to $k \sqrt{2}$. Why does $\sin A$ then equal $1/\sqrt{2}$, not simply $1$?
because $\sin A$ is always smaller than $\tan A$, whatever the triangle
$\sin A$ is opposite over hypotenuse, $k/(k \sqrt{2})$
because $1$ is reserved as the value of $\tan A$ only, never $\sin A$
because the triangle is not really a right triangle when both legs are equal
Check your answer
because $\sin A$ is always smaller than $\tan A$, whatever the triangle — There is no such general rule — $\sin A = 1/\sqrt{2}$ specifically because it divides the opposite leg by the hypotenuse $k \sqrt{2}$, not because of a size comparison with $\tan A$.
✓ $\sin A$ is opposite over hypotenuse, $k/(k \sqrt{2})$ — (B) $\sin A$ divides the opposite leg $k$ by the hypotenuse $k \sqrt{2}$, giving $1/\sqrt{2}$, not $1$.
because $1$ is reserved as the value of $\tan A$ only, never $\sin A$ — No ratio’s value is reserved for another — $\sin A$ simply happens to equal $1/\sqrt{2}$ here because of what it divides, the opposite leg by the hypotenuse.
because the triangle is not really a right triangle when both legs are equal — A right triangle with two equal legs is still a right triangle — the right angle is unaffected by the legs being equal.
For the same triangle, $2 \cdot \sin A \cdot \cos A = 2 \cdot (1/\sqrt{2}) \cdot (1/\sqrt{2})$. What is the value?
$2$
$1/2$
$1$
$\sqrt{2}$
Check your answer
$2$ — $(1/\sqrt{2}) \cdot (1/\sqrt{2}) = 1/2$, and $2 \cdot 1/2 = 1$ — both factors of $1/\sqrt{2}$ must be multiplied together first.
$1/2$ — $\sin A \cdot \cos A = 1/2$ alone, but the expression asks for $2$ times that product, which is $1$.
$\sqrt{2}$ — $(1/\sqrt{2}) \cdot (1/\sqrt{2}) = 1/2$ has no square root left in it — multiplying by $2$ gives the whole-number answer $1$, not $\sqrt{2}$.
$\sin^2 A + \cos^2 A = 1$ This is the Pythagorean identity, proved earlier.
$\cos A = \sqrt{1 - \sin^2 A}$ We rearrange for $\cos A$ and take the positive root, since $A$ is acute.
$\tan A = (\sin A)/(\cos A) = (\sin A)/\sqrt{1 - \sin^2 A}$ $\tan A$ is $\sin A$ over $\cos A$. We substitute the expression just found for $\cos A$.
for $\sin A = 3/5$, $\cos A = \sqrt{1 - 9/25} = 4/5$ and $\tan A = (3/5)/(4/5) = 3/4$ *Check by squaring $\sin A$ and $\cos A$ and adding them.* $(3/5)^2 + (4/5)^2 = 9/25 + 16/25 = 1$. This confirms the formula for $\cos A$ is right.
Write cosine and tangent using only sine.
Start from the Pythagorean identity $\sin^2 A + \cos^2 A = 1$ holds for every acute angle.
Solve for cosine Rearrange to $\cos A = \sqrt{1 - \sin^2 A}$, taking the positive root since $A$ is acute.
Build tangent from sine and cosine $\tan A$ is $\sin A$ over $\cos A$, so substitute the expression just found for $\cos A$.
Try it with a number For $\sin A = 3/5$, $\cos A = 4/5$ and $\tan A = 3/4$. Use your own value of $\sin A$ the same way.
Check yourself
Expressing $\cos A$ using only $\sin A$, for an acute angle $A$, gives
$\cos A = 1 - \sin A$
$\cos A = \sqrt{1-\sin^2 A}$
$\cos A = \sqrt{1+\sin^2 A}$
$\cos A = 1/\sin A$
Check your answer
$\cos A = 1 - \sin A$ — The identity needs $\sin^2 A$, not $\sin A$ itself: $\cos A = \sqrt{1-\sin^2 A}$, not a direct subtraction.
✓ $\cos A = \sqrt{1-\sin^2 A}$ — (B) $\cos A = \sqrt{1-\sin^2 A}$, the positive root, since $A$ is acute.
$\cos A = \sqrt{1+\sin^2 A}$ — The Pythagorean identity is $\sin^2 A + \cos^2 A=1$, so $\cos^2 A = 1 - \sin^2 A$, with a minus, not a plus.
$\cos A = 1/\sin A$ — $1/\sin A$ is $\text{cosec} A$, not $\cos A$ — $\cos A$ is found from the Pythagorean identity, $\sqrt{1-\sin^2 A}$.
Once $\cos A = \sqrt{1-\sin^2 A}$ is known, why does $\tan A = (\sin A)/\sqrt{1-\sin^2 A}$ follow immediately, without any new proof?
because $\tan A$ and $\cos A$ happen to be numerically equal for every angle
because $\tan A$ is already $(\sin A)/(\cos A)$ by definition
because a fresh proof using Pythagoras’ theorem is required for $\tan A$ as well
because $\tan A$ no longer depends on $\cos A$ once $\sin A$ is known
Check your answer
because $\tan A$ and $\cos A$ happen to be numerically equal for every angle — $\tan A$ and $\cos A$ are not generally equal — $\tan A$ follows because it is defined as $\sin A$ over $\cos A$, and $\cos A$’s new form is simply substituted in.
✓ because $\tan A$ is already $(\sin A)/(\cos A)$ by definition — (B) Substituting the rewritten $\cos A$ into the existing definition $\tan A=(\sin A)/(\cos A)$ gives the result directly.
because a fresh proof using Pythagoras’ theorem is required for $\tan A$ as well — No new proof is needed — substituting the rewritten $\cos A$ into the existing definition $\tan A=(\sin A)/(\cos A)$ gives the result directly.
because $\tan A$ no longer depends on $\cos A$ once $\sin A$ is known — $\tan A$ still depends on $\cos A$ — it is just that $\cos A$ has been rewritten in terms of $\sin A$ and substituted in.
Using $\cos A = \sqrt{1-\sin^2 A}$ and $\tan A=(\sin A)/\cos A$, find $\tan A$ when $\sin A = 7/25$.
$7/25$
$24/25$
$24/7$
$7/24$
Check your answer
$7/25$ — $7/25$ is the given $\sin A$; $\tan A$ still needs computing using $\cos A=\sqrt{1-(7/25)^2}=24/25$, giving $7/24$.
$24/25$ — $24/25$ is $\cos A$; $\tan A$ is $\sin A$ divided by that, $7/24$.
$24/7$ — $24/7$ is $\cot A$, the reciprocal of $\tan A$; $\tan A$ itself is $7/24$.
✓ $7/24$ — (D) $\cos A = \sqrt{1-(7/25)^2}=24/25$, so $\tan A = (7/25)/(24/25)=7/24$.
At 45 degrees the two legs match, which is why sine and cosine are equal there and tangent is 1.
Worked example
Proving sec A (1 - sin A)(sec A + tan A) = 1 without circularity
start from the left side alone: $\sec A (1 - \sin A)(\sec A + \tan A)$ A non-circular proof works from one side only. We never touch the right side until the very end. That is the discipline that stops a proof from assuming what it sets out to show.
$= (1/(\cos A)) \cdot (1 - \sin A) \cdot (1/(\cos A) + (\sin A)/(\cos A))$ We rewrite $\sec A$ and $\tan A$ in terms of $\sin A$ and $\cos A$.
$= (1/(\cos A)) \cdot (1 - \sin A) \cdot ((1 + \sin A)/(\cos A))$ We combine the two terms inside the bracket, over the common denominator $\cos A$.
$= ((1 - \sin A)(1 + \sin A))/(\cos^2 A)$ We multiply the three factors together, collecting both $\cos A$ terms into $\cos^2 A$.
$= (1 - \sin^2 A)/(\cos^2 A)$ $(1 - \sin A)(1 + \sin A) = 1 - \sin^2 A$, a difference of squares.
$= (\cos^2 A)/(\cos^2 A) = 1$ *We use $\sin^2 A + \cos^2 A = 1$ once, at the very end, rearranged as $1 - \sin^2 A = \cos^2 A$.* That is what turns the fraction into $1$, the right side.
at $A = 45^\circ$: $\sec 45^\circ (1 - \sin 45^\circ)(\sec 45^\circ + \tan 45^\circ)$ multiplies out to $1$ Check the identity at one actual angle. Pick $A = 45^\circ$ from the standard-angle table. The whole expression collapses to $1$, exactly as the proof promised for every acute angle.
Prove an identity by reducing one side only.
Pick one side to work from Choose the more complicated side of the identity, and leave the other side untouched until the final line.
Rewrite in sine and cosine Replace every reciprocal or quotient ratio, such as $\sec A$, $\tan A$ or $\cot A$, with its definition in $\sin A$ and $\cos A$.
Combine into one fraction Put the terms over a common denominator, and multiply out the brackets.
Simplify using the Pythagorean identity Use $\sin^2 A + \cos^2 A = 1$, rearranged if needed, exactly once, to cancel or combine terms.
Confirm it matches the other side The working should now read exactly as the side left untouched. If it does not, look for an unused simplification.
Check yourself
To prove $\sec A(1-\sin A)(\sec A+\tan A)=1$, the worked method starts by
assuming the identity is true and checking it numerically for $A=30^\circ$
cross-multiplying both sides of the equation
rewriting the left side entirely in terms of $\sin A$ and $\cos A$
squaring both sides of the identity first
Check your answer
assuming the identity is true and checking it numerically for $A=30^\circ$ — Checking one angle only confirms that one case — the worked method proves the identity for every acute angle by rewriting the left side algebraically.
cross-multiplying both sides of the equation — The proof works from the left side alone, never manipulating both sides together — that would assume the very equality being proved.
✓ rewriting the left side entirely in terms of $\sin A$ and $\cos A$ — (C) The method starts by rewriting the left side entirely in terms of $\sin A$ and $\cos A$.
squaring both sides of the identity first — No squaring step is used — the left side is simply rewritten in terms of $\sin A$ and $\cos A$ and simplified down to $1$.
In the proof of $\sec A(1-\sin A)(\sec A+\tan A)=1$, the step $((1-\sin A)(1+\sin A))/(\cos^2 A)$ simplifies to $(1-\sin^2 A)/(\cos^2 A)$. Which fact justifies this step?
$\sin^2 A + \cos^2 A = 1$, applied at this exact step
$\tan A = (\sin A)/(\cos A)$, applied at this exact step
the difference-of-squares pattern
$\sec A = 1/(\cos A)$, applied at this exact step
Check your answer
$\sin^2 A + \cos^2 A = 1$, applied at this exact step — The Pythagorean identity is used at the next step, to turn $1-\sin^2 A$ into $\cos^2 A$ — this particular step is just the difference-of-squares expansion, $(1-\sin A)(1+\sin A)=1-\sin^2 A$.
$\tan A = (\sin A)/(\cos A)$, applied at this exact step — The $\tan A$ definition was already used earlier to rewrite the left side — this specific step is the algebraic expansion $(1-\sin A)(1+\sin A)=1-\sin^2 A$.
✓ the difference-of-squares pattern — (C) This step is the algebraic expansion $(1-\sin A)(1+\sin A) = 1-\sin^2 A$, the difference-of-squares pattern.
$\sec A = 1/(\cos A)$, applied at this exact step — The $\sec A$ definition was already used to rewrite the left side earlier — this step is simply the algebraic expansion of $(1-\sin A)(1+\sin A)$.
The proof of $\sec A(1-\sin A)(\sec A+\tan A)=1$ uses the Pythagorean identity only once, right at the end, to turn $(1-\sin^2 A)/(\cos^2 A)$ into $1$. What would go wrong if the identity were used earlier, before the left side was fully rewritten in $\sin A$ and $\cos A$?
the proof would become circular, assuming the identity to prove the identity
nothing breaks, but nothing simplifies until then either
the final answer would come out different from $1$
$\sec A$ and $\tan A$ would need to be redefined
Check your answer
the proof would become circular, assuming the identity to prove the identity — Using the already-proved Pythagorean identity at any point is not circular — circularity would mean assuming the statement under proof, $\sec A(1-\sin A)(\sec A+\tan A)=1$ itself, which never happens here.
✓ nothing breaks, but nothing simplifies until then either — (B) Using the identity earlier is not wrong in itself — it simply cannot simplify anything until the expression is already written in $\sin A$ and $\cos A$.
the final answer would come out different from $1$ — The final result stays $1$ regardless of exactly when the identity is applied — reordering valid algebraic steps does not change a correct outcome, only how directly it is reached.
$\sec A$ and $\tan A$ would need to be redefined — No definitions change — $\sec A$ and $\tan A$ keep their fixed meanings throughout; only the order of applying an already-known identity is in question.
The 20-21-29 triangle checks the identity: sin A = 20/29 and cos A = 21/29, and their squares add to 1.
SIMILARITY: every ratio depends only on the angle, proved by AA similarity. A $3$-$4$-$5$ triangle and its $6$-$8$-$10$ double both give $\sin A = 3/5$.
STANDARD ANGLES: exact values at $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$ and $90^\circ$. For instance, $\sin 30^\circ = 1/2$.
PYTHAGOREAN IDENTITY: $\sin^2 A + \cos^2 A = 1$, checked on the $20$-$21$-$29$ triangle: $(20/29)^2 + (21/29)^2 = 1$.
DERIVED IDENTITIES: divide $p^2 + b^2 = h^2$ by a different squared side. Dividing by $b^2$ gives $1 + \tan^2 A = \sec^2 A$.
NON-CIRCULAR PROOF: reduce one side alone, never the identity being proved. $\sec A (1 - \sin A)(\sec A + \tan A)$ reduces to $1$ this way. No target statement appears in the middle.
Let us step back and see what carried everything so far. Every ratio traces back to one
right-triangle picture. Each ratio depends only on the angle, never on the size of the triangle.
Similarity proves this; it is not merely asserted.
The standard angles $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$ and $90^\circ$ give exact
values worth knowing outright. $\sin^2 A + \cos^2 A = 1$, and its two derived identities, let any
ratio be rewritten. Any one ratio converts into any other this way.
Reducing one side alone, and never assuming the identity being proved, is what makes both proofs trustworthy. The ratios-well-defined proof never assumed
the ratios matched before showing it. The Pythagorean identity was built once, from Pythagoras’
theorem alone, then reused everywhere else.
Pick any acute angle you like: every identity above still holds for it. You can rebuild any
ratio you forget from these two facts alone.
Check yourself
Every ratio in this chapter is built from
a separate triangle for each of the six ratios
a circle of radius $1$, as used for angles beyond $90^\circ$
measuring many different triangles and averaging the ratios found
one right triangle, with ratios fixed by the angle alone
Check your answer
a separate triangle for each of the six ratios — One right triangle supplies all six ratios at once — opposite, adjacent and hypotenuse are simply combined differently each time.
a circle of radius $1$, as used for angles beyond $90^\circ$ — This chapter’s ratios come from a right triangle, not a circle — the circle picture belongs to angles beyond the acute range, outside this chapter’s scope.
measuring many different triangles and averaging the ratios found — No averaging is needed — the proof already shows the ratio is exactly fixed by the angle, in any one triangle carrying it.
✓ one right triangle, with ratios fixed by the angle alone — (D) One right triangle picture supplies every ratio, all fixed by the angle alone, never by the triangle’s size.
This chapter’s identities all trace back to one relationship. Which one?
Pythagoras’ theorem, $p^2+b^2=h^2$, divided by different sides’ squares
the standard-angle table’s five exact values
the AA similarity criterion from Chapter 6
the definitions of $\sin A$, $\cos A$ and $\tan A$ alone, with no further theorem needed
Check your answer
✓ Pythagoras’ theorem, $p^2+b^2=h^2$, divided by different sides’ squares — (A) Every identity in this chapter comes from Pythagoras’ theorem, divided by a different side’s square each time.
the standard-angle table’s five exact values — The standard-angle table lists specific values at five angles; the identities, $\sin^2 A+\cos^2 A=1$ and its two derived forms, hold for every acute angle and come from Pythagoras’ theorem.
the AA similarity criterion from Chapter 6 — Similarity proves the ratios depend only on the angle; the identities linking $\sin A$, $\cos A$ and the rest come from Pythagoras’ theorem, a separate fact.
the definitions of $\sin A$, $\cos A$ and $\tan A$ alone, with no further theorem needed — The definitions alone give the ratios their names — proving that $\sin^2 A + \cos^2 A=1$ needs an extra step, dividing Pythagoras’ theorem by $h^2$.
A later course introduces angles beyond $90^\circ$ using a unit circle instead of a right triangle. What carries over unchanged from this chapter?
the six named ratios and the identity $\sin^2 A+\cos^2 A=1$
the right-triangle picture itself, extended to angles beyond $90^\circ$
the standard-angle table’s exact values, unchanged for every new angle
the proof that ratios depend only on the angle, unchanged in every detail
Check your answer
✓ the six named ratios and the identity $\sin^2 A+\cos^2 A=1$ — (A) The six named ratios and the identity $\sin^2 A+\cos^2 A=1$ carry over unchanged; only the picture producing them changes.
the right-triangle picture itself, extended to angles beyond $90^\circ$ — A right triangle cannot contain an angle of $90^\circ$ or more inside it — the unit-circle picture replaces the triangle for those angles; the ratios and the identity are what carry over.
the standard-angle table’s exact values, unchanged for every new angle — The standard table lists only five specific angles — new angles beyond $90^\circ$ need their own values; what carries over is the ratios’ names and the identity linking them.
the proof that ratios depend only on the angle, unchanged in every detail — The similarity-based proof is specific to a right-triangle picture; for angles beyond $90^\circ$ a different (unit-circle) argument takes over — the ratios and the core identity are what stays the same.
Check yourself: the whole chapter
A right triangle has an acute angle $A$. If the triangle is scaled up so every side becomes exactly twice as long, what happens to $\sin A$?
$\sin A$ also doubles, since the side opposite $A$ doubles in length. Doubling the triangle doubles every ratio derived from it, by this reasoning.
$\sin A$ becomes half its original value, since the hypotenuse grows faster than the opposite side.
$\sin A$ becomes undefined, since a ratio is only meaningful for the triangle it was first measured in.
$\sin A$ stays exactly the same, since it depends only on the angle $A$, not on the size of the triangle.
Check your answer
$\sin A$ also doubles, since the side opposite $A$ doubles in length. Doubling the triangle doubles every ratio derived from it, by this reasoning. — The opposite side does double, but so does the hypotenuse — since both the numerator and the denominator of $\sin A$ double together, the ratio itself is unchanged.
$\sin A$ becomes half its original value, since the hypotenuse grows faster than the opposite side. — Scaling a triangle stretches every side by the exact same factor — there is no way for the hypotenuse to grow “faster” than any other side of the same triangle.
$\sin A$ becomes undefined, since a ratio is only meaningful for the triangle it was first measured in. — A trigonometric ratio is defined precisely so that it does NOT depend on which triangle carries the angle — that is the whole reason it is useful.
✓ $\sin A$ stays exactly the same, since it depends only on the angle $A$, not on the size of the triangle. — (D) $\sin A$, like every trigonometric ratio, depends only on the angle A itself — scaling the triangle up or down never changes it.
Using the standard-angle table, what is $\cos 60^\circ$?
$\sqrt{3}/2$
$1/\sqrt{2}$
$1/2$
$\sqrt{3}$
Check your answer
$\sqrt{3}/2$ — $\sqrt{3}/2$ is the value of $\sin 60^\circ$ (it also equals $\cos 30^\circ$) — but $\cos 60^\circ$ itself is a different entry in the table, $1/2$.
$1/\sqrt{2}$ — $1/\sqrt{2}$ is $\cos 45^\circ$, one column over in the table — $\cos 60^\circ$ is the next entry along, $1/2$.
$\sqrt{3}$ — $\sqrt{3}$ is $\tan 60^\circ$, a completely different ratio — the question asks for $\cos$, not $\tan$.
Two acute angles, $A$ and $B$, are found to have exactly the same value of $\sin$. What can you conclude?
Nothing can be concluded — many different acute angles can share the same sine value.
$A$ and $B$ must be equal, since $\sin$ rises steadily from $0$ to $1$ as the angle runs from $0^\circ$ to $90^\circ$, so each value occurs at exactly one acute angle.
$A$ and $B$ must add up to $180^\circ$, since sine values are shared by supplementary angle pairs.
$A$ must be exactly double $B$, or $B$ exactly double $A$, since sine values scale that way between related angles. This kind of scaling shows up between several pairs of related angles in this chapter.
Check your answer
Nothing can be concluded — many different acute angles can share the same sine value. — sin rises steadily, never doubling back, as the angle runs from 0 degree to 90 degree — so within that acute range, each sine value belongs to exactly one angle, not many.
✓ $A$ and $B$ must be equal, since $\sin$ rises steadily from $0$ to $1$ as the angle runs from $0^\circ$ to $90^\circ$, so each value occurs at exactly one acute angle. — (B) sin climbs steadily from 0 to 1 across the acute angles — so equal sine values, in that range, mean equal angles.
$A$ and $B$ must add up to $180^\circ$, since sine values are shared by supplementary angle pairs. — Supplementary angles are outside the range this chapter works in — every angle here is acute, strictly between 0 degree and 90 degree, so this rule has no bearing on the question.
$A$ must be exactly double $B$, or $B$ exactly double $A$, since sine values scale that way between related angles. This kind of scaling shows up between several pairs of related angles in this chapter. — Nothing about sharing a sine value makes one angle a multiple of the other — the only thing it guarantees, for two acute angles, is that the angles are equal.
For an acute angle $A$, $\sin A = 3/5$. Using the identity $\sin^2 A + \cos^2 A = 1$, what is $\cos A$?
$2/5$
$-4/5$
$4/5$
$\sqrt{34}/5$
Check your answer
$2/5$ — The identity needs $\sin A$ SQUARED before subtracting — $1 - (3/5)^2 = 1 - 9/25 = 16/25$, not $1 - 3/5$.
$-4/5$ — $\sqrt{16/25} = 4/5$ is the value taken — but every ratio of an acute angle is positive, so the negative root is not a valid answer here.
✓ $4/5$ — (C) $\cos^2 A = 1 - (3/5)^2 = 1 - 9/25 = 16/25$, so $\cos A = 4/5$, the positive root since A is acute.
$\sqrt{34}/5$ — The identity is $\sin^2 A + \cos^2 A = 1$, so $\cos^2 A = 1 - \sin^2 A$ — subtracting, not adding, $\sin^2 A$ to find $\cos^2 A$.
Which of these identities is correctly derived from Pythagoras’ theorem, the same way $\sin^2 A + \cos^2 A = 1$ is?
$1 - \tan^2 A = \sec^2 A$
$\tan^2 A + \sec^2 A = 1$
$1 + \cot^2 A = \sec^2 A$
$1 + \tan^2 A = \sec^2 A$
Check your answer
$1 - \tan^2 A = \sec^2 A$ — Dividing Pythagoras’ theorem by the adjacent side’s square gives $1 + \tan^2 A = \sec^2 A$, with a plus sign, not a minus — flipping the sign breaks the identity.
$\tan^2 A + \sec^2 A = 1$ — $\tan A$ and $\sec A$ are not bounded between 0 and 1 the way $\sin A$ and $\cos A$ are, so there is no reason for their squares to sum to 1 — this only copies the earlier identity’s shape, not its actual derivation.
$1 + \cot^2 A = \sec^2 A$ — Dividing by the adjacent side’s square gives tan paired with sec; dividing by the opposite side’s square gives cot paired with cosec — mixing cot with sec crosses the two separate results.
✓ $1 + \tan^2 A = \sec^2 A$ — (D) Dividing Pythagoras’ theorem by the adjacent side’s square gives $1 + \tan^2 A = \sec^2 A$, the same way dividing by the hypotenuse’s square gives $\sin^2 A + \cos^2 A = 1$.
A tower’s height comes from a 50 m distance and a 30 degree angle, giving 50 tan 30, about 28.87 m.
You may never draw a right triangle again after Class 10. But sin, cos and tan describe slopes, heights and corners you meet often. Here are seven of those places.
Finding a water tower’s height. You stand $50$ m from the foot of a water tower. From there, its top is at an angle of elevation of $30^\circ$. The maths: the height is $50$ times $\tan 30^\circ$, read straight off the standard-angle table. $\tan 30^\circ = 1/\sqrt{3}$. So the height is $50 \cdot \tan 30^\circ = 50/\sqrt{3} \approx 28.87$ m.
Designing a wheelchair ramp. A shop is building a wheelchair ramp. The rule is $1$ m of rise for every $12$ m of ground it covers. The maths: the tangent of the ramp’s angle is $1/12$, since $\tan A$ is the rise over the run. That angle is about $4.8^\circ$, a gentle ramp.
A steep ghat road. A hill road’s sign warns of a slope of $1$ in $5$. The road rises $1$ m for every $5$ m of level ground. The maths: the road’s angle is the one whose tangent is $1/5$. As a slope nears vertical, $\tan$ grows with no limit, unlike $\sin$ and $\cos$, which never pass $1$. A “$1$ in $5$” slope makes an angle of about $11.3^\circ$.
Two triangles share a rise of 1 over runs of 12 and 5, giving foot angles of 4.8 and 11.3 degrees.
The shelf frame’s diagonal: 90 squared plus 120 squared is 22500, and so is 150 squared, so 150 cm proves the corner square.
Leaning a ladder to paint a wall. A $5$ m ladder leans against a wall, with its foot $3$ m out. It makes an angle $A$ with the ground. The maths: the ground, the wall and the ladder make a right triangle. So $\cos A$ is the ground side over the ladder. $\cos A = 3/5$. The wall side is $\sqrt{5^2 - 3^2} = 4$ m, so $\sin A = 4/5$ and the ladder reaches $4$ m up.
Checking a shop shelf frame is square. You build a shelf frame $90$ cm along one side and $120$ cm along the other. You want a true right angle at the corner. The maths: the corner is square only when the three sides fit Pythagoras’ theorem, $a^2 + b^2 = c^2$. $90^2 + 120^2 = 8100 + 14400 = 22500$, and $150^2 = 22500$ too. So a $150$ cm diagonal proves the frame square.
Cutting a diamond kite. You are cutting a diamond kite. Where the spine meets the cross-stick, one triangle has a $45^\circ$ corner. The maths: since $\tan 45^\circ = 1$, the two straight edges of that triangle are equal. If one edge is $21$ cm, so is the other, and the slanted edge is $21 \sqrt{2} \approx 29.70$ cm.
Cutting cloth from a design sheet. Your shop’s banner design sheet gives $\sin A = 3/5$ for one corner of a triangular flag. The maths: you get $\cos A$ from $\sin^2 A + \cos^2 A = 1$, without measuring the corner again. $\cos A = \sqrt{1 - (3/5)^2} = \sqrt{16/25} = 4/5$.
Your turn. A tailor’s pattern for a triangular pocket gives $\cos A = 12/13$ at one corner. What are $\sin A$ and $\tan A$? Answer: $\sin A = \sqrt{1 - (12/13)^2} = \sqrt{25/169} = 5/13$. So $\tan A = (5/13)/(12/13) = 5/12$.
practice Given $\tan A = 12/35$ for an acute angle $A$, find $\sin A$ and $\cos A$. (Worked in full below — read it, then do the next two the same way.)
practice In a right triangle the hypotenuse is $53$ cm and one leg is $45$ cm. Find $\sin A$ and $\cos A$ for the angle $A$ opposite the other leg, and check your two answers with $\sin^2 A + \cos^2 A = 1$. (Start the same way, but here two sides are given, so Pythagoras’ theorem gives the third straight away and there is no $k$ to carry.)
practice Given $\text{cosec} A = 61/11$ for an acute angle $A$, find $\cos A$ and $\cot A$. (Same first move — $\text{cosec}$ is hypotenuse over opposite, so name those two sides first and let Pythagoras’ theorem give the third.)
practice Given $\cos A = 5/13$ for an acute angle $A$, find every other ratio.
practice In a right triangle, the two legs are $8$ and $15$. Find all six ratios of the angle opposite the shorter leg.
practice Two acute angles $P$ and $Q$ satisfy $\sin P = \sin Q$. What can you conclude about $P$ and $Q$, and why?
practice Evaluate $(\sin A)/(\cos A) - \tan A$ for any acute angle $A$.
practice State whether each is true or false, with a reason: (a) $\sin A$ can equal $1.2$ for some acute angle $A$. (b) $\sec A$ is always at least $1$.
practice State whether each is true or false, with a reason: (a) $\sin A$ is the product of $\sin$ and $A$. (b) $\tan A$ has no upper bound as $A$ approaches $90^\circ$.
Answers
$\sin A = 12/37$ and $\cos A = 35/37$.
The other leg is $\sqrt{53^2 - 45^2} = \sqrt{2809 - 2025} = \sqrt{784} = 28$ cm, so $\sin A = 28/53$ and $\cos A = 45/53$. The check gives $(784 + 2025)/2809 = 1$.
The adjacent side is $\sqrt{61^2 - 11^2} = \sqrt{3721 - 121} = \sqrt{3600} = 60$, so $\cos A = 60/61$ and $\cot A = 60/11$.
$\sin A = 12/13$, $\tan A = 12/5$, $\text{cosec} A = 13/12$, $\sec A = 13/5$, $\cot A = 5/12$.
hypotenuse $= \sqrt{8^2 + 15^2} = 17$; for the angle opposite the leg $8$: $\sin A = 8/17$, $\cos A = 15/17$, $\tan A = 8/15$, $\text{cosec} A = 17/8$, $\sec A = 17/15$, $\cot A = 15/8$.
$P = Q$ — $\sin$ rises without repeating any value as an acute angle runs from $0^\circ$ to $90^\circ$, so equal sines force equal angles.
$0$ — $\tan A$ is defined as $(\sin A)/(\cos A)$, so the two terms cancel exactly.
(a) False — $\sin A$ can never exceed $1$. (b) True — $\sec A = 1/(\cos A)$ and $\cos A \leq 1$.
(a) False — $\sin A$ is one inseparable symbol, not a product. (b) True — $\tan A$ has no upper bound as $A$ approaches $90^\circ$.
Exercise 8.1 — further practice
practice Given $\sin A = 7/25$ for an acute angle $A$, find $\cos A$ and $\tan A$.
practice In a right triangle $A B C$, right-angled at $B$, $\tan A = 9/40$. Find $\sin A$ and $\cos A$.
practice In a right triangle $\text{PQR}$, right-angled at $Q$, the legs are $\text{PQ} = 7$ cm and $\text{QR} = 24$ cm. Find $\sin R$, $\cos R$ and $\tan R$.
practice Given $\cot A = 2/3$ for an acute angle $A$, find $\sin A$ and $\cos A$.
practice Two acute angles $X$ and $Y$ satisfy $\cos X = \cos Y$. What can you conclude about $X$ and $Y$, and why?
practice Simplify $\sec A \cdot \sin A \cdot \cot A$ for any acute angle $A$.
practice State whether true or false, with a reason: (a) $\cot A$ can equal $0$ for some acute angle $A$. (b) $\sin A$ is never negative for an acute angle $A$.
practice For an acute angle $A$, $\sin A = 2 k$ and $\cos A = k$ for some positive number $k$. Find $k$.
Answers
$\cos A = 24/25$, $\tan A = 7/24$.
$\sin A = 9/41$, $\cos A = 40/41$.
$\sin R = 7/25$, $\cos R = 24/25$, $\tan R = 7/24$.
$\sin A = 3/\sqrt{13}$, $\cos A = 2/\sqrt{13}$.
$X = Y$ — $\cos$ falls without repeating any value as an acute angle runs from $0^\circ$ to $90^\circ$, so equal cosines force equal angles.
$1$.
(a) False — $\cot A = (\cos A)/(\sin A)$, and for an acute angle neither $\cos A$ nor $\sin A$ is $0$, so $\cot A$ is never $0$. (b) True — $\sin A$ is a ratio of positive side lengths, always positive for an acute angle.
practice Evaluate $2 \cdot \sin 30^\circ \cdot \cos 30^\circ$. (Worked in full below — read it, then do the next two the same way.)
practice Evaluate $\sin^2 30^\circ + \cos^2 60^\circ + \tan^2 45^\circ$. (Start the same way — write the three table values down before squaring anything.)
practice In triangle $\text{PQR}$, right-angled at $Q$, angle $P = 45^\circ$ and $\text{PQ} = 7$ cm. Find $\text{QR}$ and $\text{PR}$. (The table used the other way round — pick the ratio that joins the side you have to the side you want, then substitute the table value.)
practice Choose the correct value of $\text{cosec} 30^\circ$: (a) $1$ (b) $2$ (c) $\sqrt{2}$ (d) $2/\sqrt{3}$.
practice State whether true or false, with a reason: (a) $\tan A$ increases as acute angle $A$ increases from $0^\circ$ to $90^\circ$. (b) $\sec 0^\circ$ equals $0$.
practice In a right triangle $A B C$, right-angled at $B$, angle $A = 60^\circ$ and side $B C = 6 \sqrt{3}$ cm. Find $A B$ and $A C$.
Answers
$1/2$.
$1$.
$-1/4$.
$2$.
(b) $2$.
(a) True — $\tan A = (\sin A)/(\cos A)$, and $\sin A$ rises while $\cos A$ falls as $A$ runs from $0^\circ$ to $90^\circ$, so the ratio only grows. (b) False — $\sec 0^\circ = 1/(\cos 0^\circ) = 1/1 = 1$, not $0$.
practice If $\cos A = 12/13$ for an acute angle $A$, find $\sin A$ and $\tan A$ using the Pythagorean identity, without drawing a triangle. (Worked in full below — read it, then do the next two the same way.)
practice Express $\cot A$ in terms of $\sin A$ alone, for an acute angle $A$. (Start the same way — the identity gives $\cos A$ first, then the quotient relation turns it into $\cot A$.)
practice Prove $(1 - \cos A)(1 + \cos A) = \sin^2 A$, working from the left side only. (Multiply the two brackets out first, then use the Pythagorean identity once. The right side is never touched.)
practice Express $\sec A$ in terms of $\tan A$ alone.
practice If $\cos A = 3/5$, find $\sin A$ and $\tan A$ using the Pythagorean identity, without drawing a triangle.
practice Prove $(\sin A)/(1 + \cos A) + (1 + \cos A)/(\sin A) = 2 \cdot \text{cosec} A$, working from the left side only.
practice Prove $(1 - \tan^2 A)/(1 + \tan^2 A) = \cos^2 A - \sin^2 A$, working from the left side only.
Answers
$\sin A = 5/13$ and $\tan A = 5/12$.
$\cos A = \sqrt{1 - \sin^2 A}$, so $\cot A = (\cos A)/(\sin A) = \sqrt{1 - \sin^2 A}/(\sin A)$.
The left side multiplies out to $1 - \cos^2 A$. The identity $\sin^2 A + \cos^2 A = 1$ rearranges to $1 - \cos^2 A = \sin^2 A$, which is the right side. The right side was never altered.
$\sec A = \sqrt{1 + \tan^2 A}$, from $1 + \tan^2 A = \sec^2 A$.
(c) $\text{cosec}^2 A$.
$\sin A = \sqrt{1 - 9/25} = 4/5$; $\tan A = (4/5)/(3/5) = 4/3$.
Combine over the common denominator $\sin A (1 + \cos A)$: the numerator becomes $\sin^2 A + (1 + \cos A)^2 = \sin^2 A + 1 + 2 \cos A + \cos^2 A = 2 + 2 \cos A$, using the identity. The fraction is $(2(1 + \cos A))/(\sin A (1 + \cos A)) = 2/(\sin A) = 2 \cdot \text{cosec} A$.
Multiply numerator and denominator by $\cos^2 A$: $(\cos^2 A - \sin^2 A)/(\cos^2 A + \sin^2 A) = (\cos^2 A - \sin^2 A)/1 = \cos^2 A - \sin^2 A$.
Exercise 8.3 — further practice
practice If $\sin A = 1/3$ for an acute angle $A$, find $\cos A$ using the Pythagorean identity.
practice If $\cos A = 2/3$ for an acute angle $A$, find $\sin A$ and $\cot A$ using the Pythagorean identity.
practice Given $\sec A = 5/4$ for an acute angle $A$, find the value of $\tan^2 A$ using the identity $1 + \tan^2 A = \sec^2 A$.
practice Given $\text{cosec} A = 13/12$ for an acute angle $A$, find the value of $\cot^2 A$ using the identity $1 + \cot^2 A = \text{cosec}^2 A$.
practice Express $\text{cosec} A$ in terms of $\cot A$ alone.
practice State whether true or false, with a reason: (a) $\sin^2 A - \cos^2 A$ can be written as $1 - 2 \cos^2 A$. (b) $\sec^2 A - \tan^2 A = 0$ for every acute angle $A$.
practice Prove $(\cot A - \cos A)/(\cot A + \cos A) = (\text{cosec} A - 1)/(\text{cosec} A + 1)$, working from the left side only.
practice Prove $(\tan A)/(1 - \cot A) + (\cot A)/(1 - \tan A) = 1 + \sec A \cdot \text{cosec} A$, working from the left side only.
Answers
$\cos A = (2 \sqrt{2})/3$.
$\sin A = \sqrt{5}/3$, $\cot A = 2/\sqrt{5}$.
$\tan^2 A = 9/16$.
$\cot^2 A = 25/144$.
$\text{cosec} A = \sqrt{1 + \cot^2 A}$, from $1 + \cot^2 A = \text{cosec}^2 A$, taking the positive root since $A$ is acute.
(a) True — since $\sin^2 A = 1 - \cos^2 A$, $\sin^2 A - \cos^2 A = 1 - \cos^2 A - \cos^2 A = 1 - 2 \cos^2 A$. (b) False — $\sec^2 A - \tan^2 A = 1$ for every acute angle $A$, not $0$.
Write $\cot A = (\cos A)/(\sin A)$. The numerator becomes $(\cos A)/(\sin A) - \cos A = (\cos A \cdot (1 - \sin A))/(\sin A)$ and the denominator becomes $(\cos A)/(\sin A) + \cos A = (\cos A \cdot (1 + \sin A))/(\sin A)$. The fraction is $(1 - \sin A)/(1 + \sin A)$, since $\cos A$ and $\sin A$ cancel. Writing $\text{cosec} A = 1/(\sin A)$ on the right side gives $(1/(\sin A) - 1)/(1/(\sin A) + 1) = (1 - \sin A)/(1 + \sin A)$, the same expression.
Write $\tan A = (\sin A)/(\cos A)$ and $\cot A = (\cos A)/(\sin A)$. The left side becomes $(\sin^2 A)/(\cos A \cdot (\sin A - \cos A)) - (\cos^2 A)/(\sin A \cdot (\sin A - \cos A)) = (\sin^3 A - \cos^3 A)/(\sin A \cdot \cos A \cdot (\sin A - \cos A))$. Since $\sin^3 A - \cos^3 A = (\sin A - \cos A) \cdot (\sin^2 A + \sin A \cdot \cos A + \cos^2 A) = (\sin A - \cos A) \cdot (1 + \sin A \cdot \cos A)$, the $(\sin A - \cos A)$ factor cancels, leaving $(1 + \sin A \cdot \cos A)/(\sin A \cdot \cos A) = 1/(\sin A \cdot \cos A) + 1 = \sec A \cdot \text{cosec} A + 1$, the right side.
