Trigonometry Class 10: Identities and How to Prove Them

Trigonometry in Class 10 (Introduction to Trigonometry) ends with identities and proofs, and this is where many students feel stuck. This post explains where the three identities come from, the ratio values you must know, and a few simple proof patterns that work again and again. Each step is shown, and quick check questions with answers come at the end.

First, the six ratios

Take a right triangle ABC, right-angled at B. For angle A: BC is the opposite side, AB is the adjacent side and AC is the hypotenuse.

Two more relations you will use all the time: tan A = sin A / cos A and cot A = cos A / sin A.

Keep in mind:

• "sin A" is one symbol. It is not "sin" multiplied by "A".
• The ratios depend only on the angle, not on the size of the triangle.
• The hypotenuse is the longest side, so sin A and cos A are never more than 1.

Standard values

How the values come about:

• 45°: an isosceles right triangle with equal sides a has hypotenuse a√2, so sin 45° = cos 45° = 1/√2 and tan 45° = 1.
• 30° and 60°: cut an equilateral triangle of side 2a in half. The half has sides a, a√3 and 2a.
• 0° and 90°: tan 90° and sec 90° are not defined (they divide by cos 90° = 0); cosec 0° and cot 0° are not defined (they divide by sin 0° = 0). But cot 90° = cos 90° / sin 90° = 0.

As the angle goes from 0° to 90°, sin A rises from 0 to 1 and cos A falls from 1 to 0.

Where the three identities come from

Start with Pythagoras in triangle ABC:

AB² + BC² = AC²

Divide by AC²: (AB/AC)² + (BC/AC)² = 1, so sin² A + cos² A = 1 (true for 0° ≤ A ≤ 90°)

Divide by AB²: 1 + (BC/AB)² = (AC/AB)², so 1 + tan² A = sec² A (needs cos A ≠ 0, so 0° ≤ A < 90°)

Divide by BC²: (AB/BC)² + 1 = (AC/BC)², so cot² A + 1 = cosec² A (needs sin A ≠ 0, so 0° < A ≤ 90°)

An identity is true for every angle at which all its parts are defined. Always check that no denominator becomes zero.

Useful rearranged forms:

• sin² A = 1 - cos² A, and cos² A = 1 - sin² A
• sec² A - tan² A = 1
• cosec² A - cot² A = 1

Using the identities: worked examples

Example 1. Find 9 sec² A - 9 tan² A.

• = 9(sec² A - tan² A)
• = 9 × 1 = 9

Example 2. Write cos A and tan A in terms of sin A (A acute).

• cos A = √(1 - sin² A)
• tan A = sin A / cos A = sin A / √(1 - sin² A)

Example 3. If sin A = 3/5 (A acute), find cos A.

• Let opposite = 3k and hypotenuse = 5k.
• Adjacent = √(25k² - 9k²) = √(16k²) = 4k
• cos A = 4k/5k = 4/5

Example 4. Evaluate sin² 30° + cos² 30°.

• = (1/2)² + (√3/2)² = 1/4 + 3/4 = 1, just as the first identity says.

Proof patterns that work

To prove an identity, start from one side (usually the more complicated one) and turn it into the other side. Do not move terms across the equals sign as if it were an equation.

Pattern 1: change everything to sin and cos.

Prove: sec A (1 - sin A)(sec A + tan A) = 1, for 0° ≤ A < 90°.

• LHS = (1/cos A)(1 - sin A)(1/cos A + sin A/cos A)
• = (1/cos A)(1 - sin A) × (1 + sin A)/cos A
• = (1 - sin A)(1 + sin A) / cos² A
• = (1 - sin² A) / cos² A
• = cos² A / cos² A = 1 = RHS

Pattern 2: use (a - b)(a + b) = a² - b².

Prove: (sec A + tan A)(1 - sin A) = cos A, for 0° ≤ A < 90°.

• LHS = (1/cos A + sin A/cos A)(1 - sin A)
• = (1 + sin A)(1 - sin A) / cos A
• = (1 - sin² A) / cos A
• = cos² A / cos A = cos A = RHS

Pattern 3: add fractions with a common denominator.

Prove: sin A / (1 + cos A) + (1 + cos A) / sin A = 2 cosec A, for 0° < A ≤ 90°.

• LHS = sin² A + (1 + cos A)² / sin A (1 + cos A)
• Numerator = sin² A + 1 + 2 cos A + cos² A = (sin² A + cos² A) + 1 + 2 cos A = 2 + 2 cos A = 2(1 + cos A)
• LHS = 2(1 + cos A) / sin A (1 + cos A) = 2 / sin A = 2 cosec A = RHS

Pattern 4: spot an identity hiding in the expression.

Prove: (1 + tan² A) cos² A = 1, for 0° ≤ A < 90°.

• LHS = sec² A × cos² A = (1/cos² A) × cos² A = 1 = RHS

Common mistakes to avoid

• Treating ratios like numbers you can split. sin(A + B) is not sin A + sin B. For A = B = 30°: sin 60° = √3/2, but sin 30° + sin 30° = 1.
• Writing sin² A as sin A². sin² A means (sin A)².
• Using cot A = 1/tan A where tan A is not defined. At 90°, use cot 90° = cos 90° / sin 90° = 0.
• Forgetting the allowed angles. 1 + tan² A = sec² A does not work at 90°.
• Working on both sides at once without showing that one side becomes the other.
• Mixing up sin 30° and sin 60°. sin 30° = 1/2 = cos 60°.

Quick check questions

1. What is sec 45°? Answer: 1/cos 45° = √2.
2. If tan A = 4/3, find sin A and cos A. Answer: Sides 4k, 3k, hypotenuse 5k, so sin A = 4/5 and cos A = 3/5.
3. Find the value of 5 cosec² A - 5 cot² A (0° < A ≤ 90°). Answer: 5 × 1 = 5.
4. Is tan 0° defined? Answer: Yes. tan 0° = 0/1 = 0.
5. Which angle has sin A = cos A? Answer: 45°, where both are 1/√2.
6. Evaluate 2 tan² 45° + cos² 30° - sin² 60°. Answer: 2(1) + 3/4 - 3/4 = 2.

Practise more identity proofs with hints at every step when you study this chapter with Joy on Learnijoy.