Trigonometric Ratios of Complementary Angles

What Are Trigonometric Ratios of Complementary Angles?

Two angles are complementary if they sum to $90°$, and the trigonometric ratios of complementary angles state that each ratio of an angle equals the corresponding co-ratio of its complement. So for any acute angle $\theta$, the angle $(90° - \theta)$ is its complement, and the two angles' ratios are linked by a fixed set of identities — also called the cofunction identities.

Here are all six, the central result of this topic:

$$\sin(90° - \theta) = \cos\theta \qquad \cos(90° - \theta) = \sin\theta$$

$$\tan(90° - \theta) = \cot\theta \qquad \cot(90° - \theta) = \tan\theta$$

$$\sec(90° - \theta) = \csc\theta \qquad \csc(90° - \theta) = \sec\theta$$

The pattern: each ratio turns into its co-ratio (sine ↔ cosine, tangent ↔ cotangent, secant ↔ cosecant) when the angle is replaced by its complement. These are the same pairings collected in the cofunction identities; here we focus on the complementary-angle logic behind them.

Complementary is not supplementary. Complementary angles sum to $90°$; supplementary angles sum to $180°$. The identities above hold only for the $90°$ pairing — mixing the two is a common early error.

How Are the Complementary-Angle Identities Proved?

The proof needs nothing beyond a single right triangle. Take a right triangle $ABC$ with the right angle at $C$. The two acute angles, at $A$ and $B$, must add to $90°$ (the angles of any triangle sum to $180°$, and $C$ already uses $90°$). So if $\angle A = \theta$, then $\angle B = 90° - \theta$ — the two acute angles are always complementary.

Now label the sides relative to $\angle A = \theta$:

• the side opposite $A$ is $BC = a$,

• the side adjacent to $A$ is $AC = b$,

• the hypotenuse is $AB = c$.

By definition, $\sin\theta = \frac{a}{c}$ and $\cos\theta = \frac{b}{c}$.

Here is the key observation: the side opposite $A$ is the side adjacent to $B$, and vice versa. So for the angle $\angle B = 90° - \theta$:

• its opposite side is $b$ (which was adjacent to $A$),

• its adjacent side is $a$ (which was opposite $A$).

Therefore:

$$\sin(90° - \theta) = \sin B = \frac{\text{opposite to }B}{\text{hypotenuse}} = \frac{b}{c} = \cos\theta$$

$$\cos(90° - \theta) = \cos B = \frac{\text{adjacent to }B}{\text{hypotenuse}} = \frac{a}{c} = \sin\theta$$

The tangent identity follows from the quotient relation:

$$\tan(90° - \theta) = \frac{\sin(90° - \theta)}{\cos(90° - \theta)} = \frac{\cos\theta}{\sin\theta} = \cot\theta$$

and the secant/cosecant identities follow by taking reciprocals. The whole family rests on one fact: swapping the two acute angles of a right triangle swaps "opposite" and "adjacent."

A Quick Numerical Check

You can verify the identities against the trigonometric ratios of specific angles you already know. Take $\theta = 30°$, so $90° - \theta = 60°$:

$$\sin(90° - 30°) = \sin 60° = \frac{\sqrt{3}}{2}, \qquad \cos 30° = \frac{\sqrt{3}}{2} \checkmark$$

$$\tan(90° - 30°) = \tan 60° = \sqrt{3}, \qquad \cot 30° = \sqrt{3} \checkmark$$

Both match. The identities are not approximations; they are exact, for every angle.

Examples of Trigonometric Ratios of Complementary Angles

Example 1

Evaluate $\dfrac{\sin 18°}{\cos 72°}$.

Notice $72° = 90° - 18°$, so $\cos 72° = \cos(90° - 18°) = \sin 18°$.

$$\frac{\sin 18°}{\cos 72°} = \frac{\sin 18°}{\sin 18°} = 1$$

Final answer: $1$.

Example 2

Evaluate $\tan 26° - \cot 64°$.

A first instinct is to subtract the angles or assume the two terms are unrelated and reach for a calculator. Test the relationship instead: $64° = 90° - 26°$, and $\cot(90° - \theta) = \tan\theta$, so $\cot 64° = \cot(90° - 26°) = \tan 26°$.

That makes the two terms identical:

$$\tan 26° - \cot 64° = \tan 26° - \tan 26° = 0$$

Final answer: $0$. The complementary relationship collapses the whole expression — no calculator needed.

Example 3

Evaluate $\cos 48° - \sin 42°$.

Since $48° = 90° - 42°$, we have $\cos 48° = \cos(90° - 42°) = \sin 42°$.

$$\cos 48° - \sin 42° = \sin 42° - \sin 42° = 0$$

Final answer: $0$.

Example 4

If $\sec 4A = \csc(A - 20°)$, where $4A$ is an acute angle, find $A$. (standard exam question)

Convert one side into the other's co-ratio. Since $\sec\theta = \csc(90° - \theta)$:

$$\sec 4A = \csc(90° - 4A)$$

So the equation becomes:

$$\csc(90° - 4A) = \csc(A - 20°)$$

The cosecants are equal, so the angles are equal:

$$90° - 4A = A - 20°$$

$$110° = 5A \implies A = 22°$$

Final answer: $A = 22°$. (Check: $4A = 88°$ is acute, as required.)

Example 5

Show that $\sin^2 35° + \sin^2 55° = 1$.

Since $55° = 90° - 35°$, we have $\sin 55° = \sin(90° - 35°) = \cos 35°$. Substitute:

$$\sin^2 35° + \sin^2 55° = \sin^2 35° + \cos^2 35°$$

By the Pythagorean identity, $\sin^2 35° + \cos^2 35° = 1$.

Final answer: the expression equals $1$.

Example 6

Evaluate $\tan 1° \cdot \tan 2° \cdot \tan 3° \cdots \tan 89°$.

Pair each angle with its complement: $\tan 1°$ with $\tan 89°$, $\tan 2°$ with $\tan 88°$, and so on. For each pair, $\tan(90° - \theta) = \cot\theta$, and $\tan\theta \cdot \cot\theta = 1$:

$$\tan 1° \cdot \tan 89° = \tan 1° \cdot \cot 1° = 1$$

Every pair from $1°$–$89°$ down to $44°$–$46°$ multiplies to $1$. The lone middle term is $\tan 45° = 1$.

$$\underbrace{(1)(1)\cdots(1)}_{44 \text{ pairs}} \times \tan 45° = 1 \times 1 = 1$$

Final answer: $1$. The whole product collapses because complementary tangents are reciprocals.

Why Complementary-Angle Ratios Earn Their Place

These identities exist because they let you trade an awkward angle for a friendlier one and cancel terms outright — which is exactly what turns a fearsome-looking expression into a one-line answer.

• They simplify before you compute. An expression like $\frac{\sin 18°}{\cos 72°}$ has no clean calculator value worth chasing — but seen as complementary angles it is just $1$. The identities are a recognition skill that saves the arithmetic entirely.

• They power a whole class of exam problems. The "evaluate $\cos 48° - \sin 42°$" and "$\tan 1° \cdots \tan 89°$" questions appear every year precisely because they reward students who see the $90°$ pairing.

• They explain the unit circle's symmetry. The reflection that swaps sine and cosine across the $45°$ line is the complementary relationship, and it is the first symmetry that makes the full trigonometric functions picture readable.

Where Students Slip on Complementary Angles

Mistake 1: Confusing complementary with supplementary

Where it slips in: Spotting the pairing in an expression.

Don't do this: Treating $\sin 30°$ and $\sin 150°$ as a complementary pair (they sum to $180°$, not $90°$).

The correct way: Complementary means the angles sum to $90°$; supplementary means $180°$. The cofunction identities only apply to the $90°$ pairing — check the sum before using $\sin(90° - \theta) = \cos\theta$. The point of confusion between these two close ideas is the near-identical names, so verifying "do these add to $90°$?" first settles it.

Mistake 2: Keeping the same ratio instead of switching to the co-ratio

Where it slips in: Applying the identity in a hurry.

Don't do this: Writing $\sin(90° - \theta) = \sin\theta$ — keeping sine instead of switching to cosine.

The correct way: The complement swaps the ratio for its co-ratio: $\sin \to \cos$, $\tan \to \cot$, $\sec \to \csc$. The first-instinct error is leaving the ratio unchanged; remembering that "the co- is the complement" forces the switch.

Mistake 3: Forgetting it is the angle, not the ratio, that is being complemented

Where it slips in: Expressions like $\cos(90° - \theta)$ buried inside a larger problem.

Don't do this: Reading $\cos(90° - \theta)$ as "$90°$ minus $\cos\theta$" — pulling the cosine outside the bracket.

The correct way: The $(90° - \theta)$ is the angle fed into the ratio; you replace the whole thing with the co-ratio of $\theta$, giving $\sin\theta$. The step that looks skippable but isn't is recognising the bracket as a single complementary angle.

Key Takeaways

• Complementary angles sum to $90°$; the two acute angles of any right triangle are complementary.

• Each ratio of an angle equals the co-ratio of its complement: $\sin(90° - \theta) = \cos\theta$, $\tan(90° - \theta) = \cot\theta$, $\sec(90° - \theta) = \csc\theta$.

• The proof rests on one fact — swapping the two acute angles swaps "opposite" and "adjacent."

• Tangents of complementary angles are reciprocals, which makes $\tan 1° \cdot \tan 2° \cdots \tan 89° = 1$.

• Complementary identities are a recognition skill: they cancel terms and replace awkward angles before any calculation.

Practice Before Moving On

Spot the $90°$ pairing in each before computing.

1. Evaluate $\dfrac{\cos 37°}{\sin 53°}$.

2. If $\tan 2A = \cot(A - 18°)$, where $2A$ is acute, find $A$.

3. Evaluate $\sec^2 20° - \cot^2 70°$.

Answer to Question 1: $\sin 53° = \sin(90° - 37°) = \cos 37°$, so the ratio is $1$.

Answer to Question 2: $\cot(A - 18°) = \tan(90° - (A - 18°)) = \tan(108° - A)$… more directly, $\tan 2A = \cot(A-18°) = \tan(90° - (A - 18°))$ gives $2A = 108° - A$, so $3A = 108°$, $A = 36°$ — re-check if you didn't land here.

Answer to Question 3: $\cot 70° = \cot(90° - 20°) = \tan 20°$, so the expression is $\sec^2 20° - \tan^2 20° = 1$.

To turn complementary-angle spotting into a reflex with a teacher, explore Bhanzu's trigonometry tutor, the high school math tutor track, or math tutoring online.

Want a Bhanzu trainer to coach your child through the cofunction identities until the $90°$ pairing jumps out instantly? Book a free demo class.

Read More

• Complementary angles — the geometry of two angles summing to $90°$.

• Trigonometric table — the standard-angle values used to check these identities.

• Sin cos tan — the primary ratios that swap with their co-ratios.

• Trigonometric identities — the wider family of identities these belong to.

• Trigonometric ratios — the six ratios these identities relate.