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Trigonometric Identities

An identity is an equation that is true for every allowed angle. From Pythagoras on a right triangle with hypotenuse 1: sin²A + cos²A = 1. Dividing by cos²A gives 1 + tan²A = sec²A (A ≠ 90°). Dividing by sin²A gives 1 + cot²A = cosec²A (A ≠ 0°). Use them to find one ratio from another and to prove other statements.

🎬 Step-by-step story

  1. Here is a right triangle whose hypotenuse is exactly 1 unit long. Angle A is at the bottom left.
  2. Because the hypotenuse is 1, the height is sin A and the base is cos A. No division needed!
  3. Now build a square on each short side. The red square has area sin²A. The blue square has area cos²A.
  4. Pythagoras says the two small squares together equal the square on the hypotenuse, whose area is 1. So sin²A + cos²A = 1.
  5. Divide the whole identity by cos²A: 1 + tan²A = sec²A. Divide by sin²A instead: 1 + cot²A = cosec²A.
  6. Free play: move the slider to change A. The squares change size, but their total is always 1.

Tip: drag the 3D scene to turn it. Use two fingers to zoom.

🤔 Common doubts, cleared

Why use a hypotenuse of exactly 1?

Then sin A = opposite/1 = opposite, so the sides themselves are sin A and cos A. It makes the picture easy.

Why is the height sin A and not cos A?

The height faces angle A, so it is the opposite side. Opposite ÷ 1 = sin A.

What does sin²A look like?

It is the area of a square whose side is sin A. The red square in the 3D.

Why is sin²A + cos²A = 1 and not sin A + cos A = 1?

Pythagoras adds the squares of the sides, not the sides. The two small squares fill the big square of area 1.

Where does 1 + tan²A = sec²A come from?

Divide every term of sin²A + cos²A = 1 by cos²A. The readout shows both sides match.

Does the identity break for some angle?

No. Move the slider to any angle: the red and blue areas change but always add to 1.

What is a trigonometric identity?

An equation like 2x = 6 is true for only one value (x = 3). An identity is true for every value where both sides make sense. For example (x + 1)² = x² + 2x + 1 is an identity. A trigonometric identity is an identity with trigonometric ratios of an angle.

Note: sin²A means (sin A)², not sin(A²).

Proof of sin²A + cos²A = 1

Take right triangle ABC with the right angle at B. By Pythagoras: AB² + BC² = AC².

Divide every term by AC²: (AB/AC)² + (BC/AC)² = 1.

But AB/AC = cos A and BC/AC = sin A. So cos²A + sin²A = 1, true for 0° ≤ A ≤ 90°.

In the 3D, the hypotenuse is 1, so the squares on the two legs have areas sin²A and cos²A and fill the unit square on the hypotenuse exactly.

The other two identities

1 + tan²A = sec²A (for 0° ≤ A < 90°): divide sin²A + cos²A = 1 by cos²A. sin²A/cos²A = tan²A, cos²A/cos²A = 1 and 1/cos²A = sec²A.

1 + cot²A = cosec²A (for 0° < A ≤ 90°): divide by sin²A instead.

Useful rearranged forms: sin²A = 1 − cos²A, cos²A = 1 − sin²A, sec²A − tan²A = 1, cosec²A − cot²A = 1. The last two can be split as (sec A − tan A)(sec A + tan A) = 1.

Using identities: one ratio gives all

If you know one ratio you can write all others. Example: express everything in terms of cot A. cosec A = √(1 + cot²A), sin A = 1/√(1 + cot²A), tan A = 1/cot A, cos A = cot A/√(1 + cot²A).

How to prove an identity (exam method)

  1. Start with the more complicated side.
  2. If stuck, write everything in sin and cos.
  3. Use the three identities and algebra (common denominator, (a − b)(a + b) = a² − b²).
  4. Stop when you reach the other side. Never move terms across the = sign as if solving an equation.

Proofs of identities are a regular 2 to 3 mark question in CBSE Class 10 board exams.

Key formulas and definitions

Worked examples

1. If sin A = 0.6, find cos A using an identity (A acute).

cos²A = 1 − sin²A = 1 − 0.36 = 0.64. cos A = 0.8.

2. Find the value of 9 sec²A − 9 tan²A.

9(sec²A − tan²A) = 9 × 1 = 9.

3. Simplify (1 + tan²θ)(1 − sin θ)(1 + sin θ).

(1 + tan²θ) = sec²θ and (1 − sin θ)(1 + sin θ) = 1 − sin²θ = cos²θ. So sec²θ × cos²θ = 1.

4. Prove that (1 − cos²A) cosec²A = 1.

1 − cos²A = sin²A. So sin²A × cosec²A = sin²A × 1/sin²A = 1.

5. Prove that cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A.

Common denominator: [cos²A + (1 + sin A)²] / [cos A(1 + sin A)] = [cos²A + 1 + 2 sin A + sin²A] / [cos A(1 + sin A)] = [2 + 2 sin A] / [cos A(1 + sin A)] = 2(1 + sin A)/[cos A(1 + sin A)] = 2/cos A = 2 sec A.

6. Prove that (sec A − tan A)² = (1 − sin A)/(1 + sin A).

Left = (1/cos A − sin A/cos A)² = (1 − sin A)²/cos²A = (1 − sin A)²/(1 − sin²A) = (1 − sin A)²/[(1 − sin A)(1 + sin A)] = (1 − sin A)/(1 + sin A).

7. If sec A + tan A = 3, find sec A − tan A and then sec A.

(sec A + tan A)(sec A − tan A) = sec²A − tan²A = 1, so sec A − tan A = 1/3. Adding: 2 sec A = 3 + 1/3 = 10/3, sec A = 5/3.

Common mistakes

Practice quiz

1. sin²A + cos²A equals:
2. 1 + tan²A equals:
3. cosec²A − cot²A equals:
4. If cos A = 5/13, sin A equals:
5. (1 − sin²A) sec²A equals:

Practice: answer these yourself

Type or choose your answer, then press Check. Use a hint if you are stuck; the full solution appears after you answer.

Frequently asked questions

Why is sin²A + cos²A always 1?

Because it is Pythagoras' theorem divided by the hypotenuse squared. The two leg squares always fill the square on the hypotenuse.

How many trigonometric identities are in Class 10?

Three: sin²A + cos²A = 1, 1 + tan²A = sec²A and 1 + cot²A = cosec²A. Everything else is built from these.

What is the best trick to prove identities?

Start from the harder side and change everything to sin and cos. Then use the three identities and simple algebra.

Where this is taught

PolandLiceum ogólnokształcące, klasa IITrigonometry
CBSE (India)Class 10Trigonometry
USA (Common Core, NGSS, AP)Grade 10Similarity, right-triangle trigonometry and proof
USA (Common Core, NGSS, AP)Grade 11Trigonometric functions

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