Trigonometric Ratios of Complementary Angles

What Are Complementary Angles?

Two angles are complementary if they add up to 90°. So $\theta$ and $(90° - \theta)$ are complementary.

In a right triangle, the two acute angles are always complementary (they sum to 90°, since the third angle is the right angle). This simple fact creates beautiful relationships between their ratios.

Key Point: If $\theta + \phi = 90°$, then $\phi = 90° - \theta$, and $\theta, \phi$ are complementary.

[Board Important] Don't confuse complementary (sum 90°) with supplementary (sum 180°). Class 10 trigonometry uses complementary.

The Complementary-Angle Relations

For an acute angle $\theta$:

$$\sin(90° - \theta) = \cos\theta, \quad \cos(90° - \theta) = \sin\theta$$ $$\tan(90° - \theta) = \cot\theta, \quad \cot(90° - \theta) = \tan\theta$$ $$\sec(90° - \theta) = \csc\theta, \quad \csc(90° - \theta) = \sec\theta$$

Why: in a right triangle, the side opposite $\theta$ is the side adjacent to $(90°-\theta)$, so 'opposite' and 'adjacent' swap — turning sin into cos, tan into cot, etc.

Key Point: Each ratio of $(90° - \theta)$ becomes its co-ratio of $\theta$: sin↔cos, tan↔cot, sec↔cosec.

[JEE/NEET Tip] The prefix 'co' in cosine, cotangent, cosecant literally means 'complement' — that's the memory hook.

Using the Relations

These relations let you rewrite a ratio of a large angle as a ratio of a smaller one, which is handy for simplification and for pairing terms.

For example, $\sin 70° = \sin(90° - 20°) = \cos 20°$. So $\dfrac{\sin 70°}{\cos 20°} = 1$.

Key Point: Look for angle pairs that sum to 90° — they collapse using the co-ratio relations.

[Board Important] Whenever you see two angles adding to 90° (like 35° and 55°, or 18° and 72°), expect a complementary-angle simplification.

Solving Equations with Complementary Angles

If an equation equates two ratios, say $\sin A = \cos B$, rewrite one side using the complement: $\cos B = \sin(90° - B)$, so $\sin A = \sin(90° - B)$, giving $A = 90° - B$, i.e. $A + B = 90°$.

Key Point: $\sin A = \cos B \Rightarrow A + B = 90°$ (for acute angles). The same logic works for tan/cot and sec/cosec pairs.

[Board Important] A very common 2-mark question: 'If $\sin 3A = \cos(A - 26°)$, find $A$.' Set $3A + (A - 26°) = 90°$.

Solved Examples

Example 1: Direct relation

Evaluate $\dfrac{\sin 35°}{\cos 55°}$.

Solution:

1. $\cos 55° = \cos(90° - 35°) = \sin 35°$.
2. Ratio $= \dfrac{\sin 35°}{\sin 35°} = 1$.

Final Answer: 1.

Takeaway: $35° + 55° = 90°$ ⇒ they are co-ratios.

Example 2: tan/cot pair

Evaluate $\tan 26° \cdot \tan 64°$.

Solution:

1. $\tan 64° = \tan(90° - 26°) = \cot 26°$.
2. $\tan 26° \cdot \cot 26° = \tan 26° \cdot \dfrac{1}{\tan 26°} = 1$.

Final Answer: 1.

Takeaway: $26° + 64° = 90°$, and $\tan\theta\cot\theta = 1$.

Example 3: Solve for A

If $\sin 3A = \cos(A - 26°)$, where $3A$ is acute, find $A$.

Solution:

1. $\cos(A - 26°) = \sin(90° - (A - 26°)) = \sin(116° - A)$.
2. So $3A = 116° - A \Rightarrow 4A = 116° \Rightarrow A = 29°$.

Final Answer: $A = 29°$.

Takeaway: Convert cos to sin, then equate angles.

Example 4: Simplify a sum

Evaluate $\cos 48° - \sin 42°$.

Solution:

1. $\sin 42° = \sin(90° - 48°) = \cos 48°$.
2. $\cos 48° - \cos 48° = 0$.

Final Answer: 0.

Takeaway: $48° + 42° = 90°$ makes the terms equal.

Example 5: sec/cosec

Evaluate $\dfrac{\sec 70°}{\csc 20°}$.

Solution:

1. $\sec 70° = \sec(90° - 20°) = \csc 20°$.
2. Ratio $= 1$.

Final Answer: 1.

Takeaway: $\sec(90° - \theta) = \csc\theta$.

Example 6: Combine with table values

Evaluate $\sin 30° \cos 60° + \cos 30° \sin 60°$ and identify the angle.

Solution:

1. $= \tfrac12\cdot\tfrac12 + \tfrac{\sqrt3}{2}\cdot\tfrac{\sqrt3}{2} = \tfrac14 + \tfrac34 = 1$.
2. This equals $\sin(30° + 60°) = \sin 90° = 1$.

Final Answer: 1.

Takeaway: Complementary angles 30° and 60° combine to 90°.

Example 7: Express in complementary form

Express $\cot 85° + \cos 75°$ in terms of ratios of angles between 0° and 45°.

Solution:

1. $\cot 85° = \cot(90° - 5°) = \tan 5°$.
2. $\cos 75° = \cos(90° - 15°) = \sin 15°$.

Final Answer: $\tan 5° + \sin 15°$.

Takeaway: Use complements to bring all angles below 45°.

Example 8: Solve a tan equation

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

Solution:

1. $\cot(A - 18°) = \tan(90° - (A - 18°)) = \tan(108° - A)$.
2. $2A = 108° - A \Rightarrow 3A = 108° \Rightarrow A = 36°$.

Final Answer: $A = 36°$.

Takeaway: Convert cot to tan, then equate angles.

Example 9: A complementary-angle ratio

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

Solution:

1. $\cos 72° = \cos(90° - 18°) = \sin 18°$.
2. $\dfrac{\sin 18°}{\sin 18°} = 1$.

Final Answer: 1.

Takeaway: $18° + 72° = 90°$ ⇒ ratio is 1.

Example 10: Mixed evaluation

Evaluate $\sin^2 25° + \sin^2 65°$.

Solution:

1. $\sin 65° = \sin(90° - 25°) = \cos 25°$, so $\sin^2 65° = \cos^2 25°$.
2. $\sin^2 25° + \cos^2 25° = 1$.

Final Answer: 1.

Takeaway: Complementary angles turn this into the master identity.