Complementary angles

Look at any right-angled triangle and a small miracle is hiding in plain sight: its two acute angles are not independent at all — they are forced to add up to exactly 90∘90^\circ. Call one of them AA and the other is automatically 90∘−A90^\circ - A, its complement. That single constraint quietly links every trigonometric ratio to a partner ratio: sine to cosine, tangent to cotangent, secant to cosecant. Learn the link once and you can rewrite sin⁡70∘\sin 70^\circ as cos⁡20∘\cos 20^\circ without a calculator, collapse ugly-looking fractions to a clean 11, or turn an equation buried in unknown angles into a simple linear one. This article builds the six complementary-angle relationships from the geometry of a single triangle, then puts them to work on the exact kind of evaluation and equation-solving questions ICSE Class 9 asks.

The core insight

For any acute angle AA, each trigonometric ratio equals the "co-ratio" of its complement 90∘−A90^\circ - A, because the same pair of triangle sides plays "opposite" for one acute angle and "adjacent" for the other.

The core idea

Why the two acute angles are complementary. In any right-angled triangle the three angles sum to 180∘180^\circ, and one of them is already 90∘90^\circ. The remaining two acute angles must therefore share the leftover 90∘90^\circ between them — if one is AA, the other is forced to be (90∘−A)(90^\circ - A).

Why the ratios swap. Relative to angle AA, one particular side is "opposite"; relative to the other acute angle (90∘−A)(90^\circ - A), that very same side is "adjacent." Sides never change, only the label depends on which angle you are looking from — and that side-relabelling is the entire reason the ratios pair up:

sin⁡A=cos⁡(90∘−A),cos⁡A=sin⁡(90∘−A)\sin A = \cos(90^\circ - A), \qquad \cos A = \sin(90^\circ - A)

where: AA is any acute angle with 0∘<A<90∘0^\circ < A < 90^\circ, and 90∘−A90^\circ - A is its complement measured in the same right triangle. The same side-swapping logic gives the other two co-pairs:

tan⁡A=cot⁡(90∘−A),cot⁡A=tan⁡(90∘−A)\tan A = \cot(90^\circ - A), \qquad \cot A = \tan(90^\circ - A)
sec⁡A=csc⁡(90∘−A),csc⁡A=sec⁡(90∘−A)\sec A = \csc(90^\circ - A), \qquad \csc A = \sec(90^\circ - A)

How to remember them. There are exactly three co-pairs — (sine, cosine), (tangent, cotangent), (secant, cosecant) — and sine is never paired with tangent or with anything outside its own pair. Crossing from one member of a pair to the other always means replacing the angle with its complement, in either direction: sin⁡A=cos⁡(90∘−A)\sin A = \cos(90^\circ - A) and, just as validly, cos⁡A=sin⁡(90∘−A)\cos A = \sin(90^\circ - A).

Figure
A right triangle XYZ right angled at Y showing side YZ as opposite to angle X but adjacent to angle Z, and side XY as adjacent to angle X but opposite to angle Z
The same side is opposite for one acute angle, adjacent for the other X Y Z A 90° − A XY YZ YZ is opposite X, XY is adjacent to X YZ is adjacent to Z, XY is opposite Z sin X = YZ over XZ = cos Z, since the same ratio has two names

Common traps

  • Pairing the wrong ratios — the three fixed co-pairs are (sine, cosine), (tangent, cotangent) and (secant, cosecant); sine is never paired with tangent.
  • Forgetting to switch the angle to its complement when switching a ratio to its co-ratio (writing sin⁡A=cos⁡A\sin A = \cos A instead of sin⁡A=cos⁡(90∘−A)\sin A = \cos(90^\circ - A)).
  • In an equation like tan⁡(3A−10∘)=cot⁡(A+40∘)\tan(3A - 10^\circ) = \cot(A + 40^\circ), equating the two given angles directly instead of first converting one side onto the same ratio as the other.
  • Applying the complementary-angle rules outside the acute range, or forgetting that the two acute angles of a right triangle must sum to exactly 90∘90^\circ, no more and no less.
Derivation AWhy sin A equals cos of its complement

The goal is to show, starting only from the two basic definitions "opposite over hypotenuse" for sine and "adjacent over hypotenuse" for cosine and the fact that a triangle's angles sum to 180∘180^\circ, that sin⁡A=cos⁡(90∘−A)\sin A = \cos(90^\circ - A) holds for every acute angle AA in a right triangle — not as a rule to memorise but as a direct consequence of how the same triangle looks from its two different acute corners.

Step 1. Fix a right-angled triangle XYZXYZ, right-angled at YY, with angle XX called AA. Since the three angles sum to 180∘180^\circ and angle YY is 90∘90^\circ, angles XX and ZZ together make up the remaining 90∘90^\circ, so angle ZZ must equal 90∘−A90^\circ - A.

Step 2. Identify the sides relative to angle XX. The side opposite XX (the one not touching XX) is YZYZ; the side adjacent to XX (touching XX, but not the hypotenuse) is XYXY; the hypotenuse is XZXZ. By definition, sin⁡X=YZXZ\sin X = \dfrac{YZ}{XZ}.

Step 3. Identify the same two sides relative to angle ZZ. Now stand at corner ZZ instead: the side opposite ZZ is XYXY, and the side adjacent to ZZ is YZYZ — the very side that was "opposite" a moment ago is now "adjacent," simply because we changed which corner we are looking from. By definition, cos⁡Z=YZXZ\cos Z = \dfrac{YZ}{XZ}.

Step 4. Compare the two ratios. Step 2 gave sin⁡X=YZ/XZ\sin X = YZ/XZ and Step 3 gave cos⁡Z=YZ/XZ\cos Z = YZ/XZ — the exact same fraction, built from the exact same pair of sides. So sin⁡X=cos⁡Z\sin X = \cos Z, and since Z=90∘−AZ = 90^\circ - A and X=AX = A, this reads sin⁡A=cos⁡(90∘−A)\sin A = \cos(90^\circ - A). Repeating the same side-relabelling argument for the remaining pairs of sides gives tan⁡A=cot⁡(90∘−A)\tan A = \cot(90^\circ - A) and sec⁡A=csc⁡(90∘−A)\sec A = \csc(90^\circ - A) in exactly the same way.

Worked example

Question. Evaluate sin⁡42∘cos⁡48∘\dfrac{\sin 42^\circ}{\cos 48^\circ} without using tables, and then find AA if tan⁡(3A−10∘)=cot⁡(A+40∘)\tan(3A - 10^\circ) = \cot(A + 40^\circ), given all angles involved are acute.

Step 1 — spot the complementary pair. In the first part, notice 42∘+48∘=90∘42^\circ + 48^\circ = 90^\circ, so 48∘=90∘−42∘48^\circ = 90^\circ - 42^\circ.

Step 2 — rewrite using the complementary-angle rule. cos⁡48∘=cos⁡(90∘−42∘)=sin⁡42∘\cos 48^\circ = \cos(90^\circ - 42^\circ) = \sin 42^\circ, so

sin⁡42∘cos⁡48∘=sin⁡42∘sin⁡42∘=1.\frac{\sin 42^\circ}{\cos 48^\circ} = \frac{\sin 42^\circ}{\sin 42^\circ} = 1.

The expression evaluates to exactly 11, with no need for the actual numerical value of sin⁡42∘\sin 42^\circ.

Step 3 — set up the equation. For the second part, use tan⁡θ=cot⁡(90∘−θ)\tan\theta = \cot(90^\circ - \theta) to rewrite the left-hand side: tan⁡(3A−10∘)=cot⁡(90∘−(3A−10∘))=cot⁡(100∘−3A)\tan(3A - 10^\circ) = \cot\big(90^\circ - (3A - 10^\circ)\big) = \cot(100^\circ - 3A). Since this now equals cot⁡(A+40∘)\cot(A + 40^\circ), the two angles inside cotangent must themselves be equal: 100∘−3A=A+40∘100^\circ - 3A = A + 40^\circ.

Step 4 — solve and check. Rearranging, 100∘−40∘=A+3A100^\circ - 40^\circ = A + 3A, so 60∘=4A60^\circ = 4A and A=15∘A = 15^\circ. Checking: with A=15∘A = 15^\circ, the left side is tan⁡35∘\tan 35^\circ and the right side is cot⁡55∘=cot⁡(90∘−35∘)=tan⁡35∘\cot 55^\circ = \cot(90^\circ - 35^\circ) = \tan 35^\circ — both sides match. The fraction equals 11, and A=15∘A = 15^\circ.

Competitive stretch

If instead you were told sin⁡(2A+10∘)=cos⁡(3A−20∘)\sin(2A + 10^\circ) = \cos(3A - 20^\circ) with all angles acute, the same method applies: rewrite cos⁡(3A−20∘)=sin⁡(90∘−(3A−20∘))=sin⁡(110∘−3A)\cos(3A-20^\circ) = \sin\big(90^\circ - (3A-20^\circ)\big) = \sin(110^\circ - 3A), equate the angles since the ratio is now matched on both sides, 2A+10∘=110∘−3A2A + 10^\circ = 110^\circ - 3A, giving 5A=100∘5A = 100^\circ and A=20∘A = 20^\circ. The pattern is always: match the ratio first by converting one side to the complement of the other, only then equate the angles.

Ask the tutor: Want to see the triangle relabelling argument worked through for the tangent-cotangent and secant-cosecant pairs as well, or try another complementary-angle equation? Ask the AI tutor to walk you through more examples.

Why does sin A equal cos of the complement instead of just being equal to cos A?

Because sine and cosine are defined using different sides — sine uses the side opposite the angle, cosine uses the side adjacent to it. For a single fixed angle A, those are genuinely different sides, so sin A and cos A are usually different numbers. What is true is that the side opposite A is the very same side that is adjacent to A's complement, which is why sin A matches cos of the complement, not cos of A itself.

Do these complementary-angle rules work for any triangle, or only right-angled ones?

They are defined and proved using a right-angled triangle, since sine, cosine and the other ratios are only defined for the acute angles of a right triangle in this syllabus. The angle A in the rules must be acute, strictly between 0° and 90°, and its complement 90° − A must also come out acute for the picture to make sense — which it always does automatically, since two positive numbers adding to 90° are both less than 90°.

How do I know whether to equate the angles directly or convert one side first?

Only equate two angles directly when they sit inside the same ratio on both sides, such as cot(100° − 3A) = cot(A + 40°). If the two sides show different ratios, like tan on one side and cot on the other, you must first use a complementary-angle rule to rewrite one ratio in terms of the other before equating angles — skipping this conversion is the single most common mistake in these equations.

Before this

Where this leads