# Trigonometric Ratios of Complementary Angles

**TL;DR**  
Two angles are complementary when they add to 90°, and the trigonometric ratio of an angle equals the co-ratio of its complement — so sin(90°−θ)=cosθ, tan(90°−θ)=cotθ, and sec(90°−θ)=cscθ. This article gives all six complementary-angle identities, proves them from a right triangle, explains why the 'co-' in cosine means complement, and works through six examples — including the classic tan 1°⋅tan 2°⋯tan 89° problem.

## 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 θ, the angle (90°−θ) 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°−θ)=cosθ  
- cos(90°−θ)=sinθ
- tan(90°−θ)=cotθ  
- cot(90°−θ)=tanθ
- sec(90°−θ)=cscθ  
- csc(90°−θ)=secθ

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](/content/math/trigonometry/cofunction-identities/index.html); 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 ∠A=θ, then ∠B=90°−θ — the two acute angles are _always complementary_.

**Now label the sides relative to ∠A=θ:**

- the side opposite A is BC=a,
- the side adjacent to A is AC=b,
- the hypotenuse is AB=c.

By definition, sinθ=
a/c and cosθ=b/c.

Here is the key observation: **the side opposite A is the side adjacent to B, and vice versa.** So for the angle ∠B=90°−θ:

- its opposite side is b (which was adjacent to A),
- its adjacent side is a (which was opposite A).

**Therefore:**

sin(90°−θ)=sinB=opposite to B/hypotenuse=b/c=cosθ

cos(90°−θ)=cosB=adjacent to B/hypotenuse=a/c=sinθ

The tangent identity follows from the quotient relation:

tan(90°−θ)=sin(90°−θ)/cos(90°−θ)=cosθ/sinθ=cotθ

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](/content/math/trigonometry/trigonometric-ratios-of-specific-angles/index.html) you already know. Take θ=30°, so 90°−θ=60°:

- sin(90°−30°)=sin(60°)=√3/2 ✓
- cos(30°)=√3/2 ✓
- tan(90°−30°)=tan(60°)=√3 ✓
- cot(30°)=√3 ✓

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

## Examples of Trigonometric Ratios of Complementary Angles

### Example 1

**Evaluate sin(18°)/cos(72°).**

Notice 72°=90°−18°, so cos(72°)=sin(18°).

Thus, sin(18°)/cos(72°)=sin(18°)/sin(18°)=1.

**Final answer:** 1.

### Example 2

**Evaluate tan(26°)−cot(64°).**

Test the relationship instead: 64°=90°−26°, and cot(90°−θ)=tanθ, so cot(64°)=tan(26°).

That makes the two terms identical:

tan(26°)−cot(64°)=tan(26°)−tan(26°)=0.

**Final answer:** 0.

### Example 3

**Evaluate cos(48°)−sin(42°).**  
Since 48°=90°−42°, we have cos(48°)=sin(42°).

Thus, 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.**

Convert one side into the other's co-ratio. Since secθ=csc(90°−θ):

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⟹A=22°.

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

### Example 5

**Show that sin(235°)+sin(255°)=1.**

Since 55°=90°−35°, we have sin(55°)=cos(35°). Substitute:
sin(235°)+sin(255°)=sin(235°)+cos(235°).

By the [Pythagorean identity](/content/math/trigonometry/pythagorean-identities/index.html), sin^2(35°)+cos^2(35°)=1.

**Final answer:** the expression equals 1.

### Example 6

**Evaluate tan(1°)⋅tan(2°)⋯tan(89°).**

Pair each angle with its complement, noting that tan(90°−θ)=cotθ. For each pair:
tan(1°)⋅tan(89°)=1, tan(2°)⋅tan(88°)=1, and so on.

The lone middle term is tan(45°)=1.

**Final answer:** 1.

## 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 sin(18°)/cos(72°) has no clean calculator value worth chasing — but seen as complementary angles it is just 1.
- **They power a whole class of exam problems.** The "evaluate cos(48°)−sin(42°)" and "tan(1°)⋯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.

## 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°−θ)=cosθ, tan(90°−θ)=cotθ, sec(90°−θ)=cscθ.
- 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°⋅tan 2°⋯tan 89°=1.
- Complementary identities are a recognition skill: they cancel terms and replace awkward angles before any calculation.

## Practice Before Moving On

1. Evaluate cos(37°)/sin(53°).
2. If tan(2A)=cot(A−18°), where 2A is acute, find A.
3. Evaluate sec(20°)−cot(70°).

**Answer to Question 1:** sin(53°)=cos(37°), so the ratio is 1.  
**Answer to Question 2:** cot(A−18°)=tan(90°−(A−18°)) gives 3A=108°, A=36°.  
**Answer to Question 3:** cot(70°)=tan(20°), so the expression is sec(20°)−tan(20°)=1.

## Frequently Asked Questions

- **What are complementary angles in trigonometry?**  
Two angles whose measures add up to 90°; the trigonometric ratios of complementary angles relate an angle to its complement in a right triangle.
- **What is the formula for sin(90°−θ)?**  
sin(90°−θ)=cosθ. 
- **Why is cos(90°−θ) equal to sinθ?**  
Because swapping the two complementary angles swaps "opposite" and "adjacent," which swaps sine and cosine.
- **What are the six complementary-angle identities?**  
- sin(90°−θ)=cosθ  
- cos(90°−θ)=sinθ  
- tan(90°−θ)=cotθ  
- cot(90°−θ)=tanθ  
- sec(90°−θ)=cscθ  
- csc(90°−θ)=secθ.  
- **How are the tangents of complementary angles related?**  
They are reciprocals: tan(90°−θ)=cotθ, so tanθ⋅tan(90°−θ)=1.
- **What is the difference between complementary and supplementary angles in trigonometry?**  
Complementary angles sum to 90°; supplementary angles sum to 180°.