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.
Evaluate $\dfrac{\cos 37°}{\sin 53°}$.
If $\tan 2A = \cot(A - 18°)$, where $2A$ is acute, find $A$.
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.
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