CBSE • Class 10 • Mathematics • Chapter 8 (Introduction to Trigonometry) • Exercise 8.3

Exercise 8.3: Introduction to Trigonometry — NCERT Solutions

Trigonometric identities — sin²θ + cos²θ = 1 and consequences.

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What this exercise covers

Pythagorean identityReciprocal identitiesIdentity-based proofs

Step-by-step solutions — Exercise 8.3

3 solved questions • Each solution includes a Socratic hint, full working and a common-mistake callout • Last reviewed 2026-09-03

Exercise 8.3 Q1 • 4 marks

Prove that (sin⁡θ+csc⁡θ)2+(cos⁡θ+sec⁡θ)2=7+tan⁡2θ+cot⁡2θ(\sin\theta + \csc\theta)^2 + (\cos\theta + \sec\theta)^2 = 7 + \tan^2\theta + \cot^2\theta.
Hint (Socratic — try this first)▾
After expanding the squares, can you use sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 and the reciprocal relations?
Step-by-step solution▾

Expand the left-hand side (LHS): (sin⁡θ+csc⁡θ)2=sin⁡2θ+2sin⁡θcsc⁡θ+csc⁡2θ(\sin\theta + \csc\theta)^2 = \sin^2\theta + 2\sin\theta\csc\theta + \csc^2\theta Since sin⁡θcsc⁡θ=1\sin\theta\csc\theta = 1: =sin⁡2θ+2+csc⁡2θ= \sin^2\theta + 2 + \csc^2\theta

Similarly: (cos⁡θ+sec⁡θ)2=cos⁡2θ+2+sec⁡2θ(\cos\theta + \sec\theta)^2 = \cos^2\theta + 2 + \sec^2\theta

Adding: LHS=(sin⁡2θ+cos⁡2θ)+4+csc⁡2θ+sec⁡2θ\text{LHS} = (\sin^2\theta + \cos^2\theta) + 4 + \csc^2\theta + \sec^2\theta =1+4+csc⁡2θ+sec⁡2θ= 1 + 4 + \csc^2\theta + \sec^2\theta

Using identities csc⁡2θ=1+cot⁡2θ\csc^2\theta = 1 + \cot^2\theta and sec⁡2θ=1+tan⁡2θ\sec^2\theta = 1 + \tan^2\theta: =5+(1+cot⁡2θ)+(1+tan⁡2θ)= 5 + (1 + \cot^2\theta) + (1 + \tan^2\theta) =7+tan⁡2θ+cot⁡2θ=RHS= 7 + \tan^2\theta + \cot^2\theta = \text{RHS}

Hence proved.

Common mistake:
Forgetting that sin⁡θ⋅csc⁡θ=1\sin\theta\cdot\csc\theta=1, so leaving the middle term unsimplified.
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Exercise 8.3 Q2 • 4 marks

Prove the identity cos⁡A1+sin⁡A+1+sin⁡Acos⁡A=2sec⁡A\dfrac{\cos A}{1 + \sin A} + \dfrac{1 + \sin A}{\cos A} = 2\sec A.
Hint (Socratic — try this first)▾
What happens if you combine the two fractions over a common denominator and use sin⁡2A+cos⁡2A=1\sin^2A + \cos^2A = 1?
Step-by-step solution▾

Take the common denominator on the LHS: cos⁡A1+sin⁡A+1+sin⁡Acos⁡A=cos⁡2A+(1+sin⁡A)2(1+sin⁡A)cos⁡A\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = \frac{\cos^2 A + (1 + \sin A)^2}{(1 + \sin A)\cos A}

Expand the numerator: cos⁡2A+1+2sin⁡A+sin⁡2A\cos^2 A + 1 + 2\sin A + \sin^2 A

Using sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1: =1+1+2sin⁡A=2+2sin⁡A=2(1+sin⁡A)= 1 + 1 + 2\sin A = 2 + 2\sin A = 2(1 + \sin A)

So: LHS=2(1+sin⁡A)(1+sin⁡A)cos⁡A=2cos⁡A=2sec⁡A=RHS\text{LHS} = \frac{2(1 + \sin A)}{(1 + \sin A)\cos A} = \frac{2}{\cos A} = 2\sec A = \text{RHS}

Hence proved.

Common mistake:
Expanding (1+sin⁡A)2(1+\sin A)^2 incorrectly as 1+sin⁡2A1 + \sin^2 A, omitting the 2sin⁡A2\sin A term.
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Exercise 8.3 Q3 • 4 marks

If sec⁡θ+tan⁡θ=p\sec\theta + \tan\theta = p, show that p2−1p2+1=sin⁡θ\dfrac{p^2 - 1}{p^2 + 1} = \sin\theta.
Hint (Socratic — try this first)▾
Can you use the identity sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1 to also find sec⁡θ−tan⁡θ\sec\theta - \tan\theta?
Step-by-step solution▾

Given sec⁡θ+tan⁡θ=p\sec\theta + \tan\theta = p.

Using the identity sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1: (sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ)=1(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1 p(sec⁡θ−tan⁡θ)=1  ⟹  sec⁡θ−tan⁡θ=1pp(\sec\theta - \tan\theta) = 1 \implies \sec\theta - \tan\theta = \frac{1}{p}

Now: p2−1=p2−1,and notep^2 - 1 = p^2 - 1, \quad \text{and note} p−1p=(sec⁡θ+tan⁡θ)−(sec⁡θ−tan⁡θ)=2tan⁡θp - \frac{1}{p} = (\sec\theta + \tan\theta) - (\sec\theta - \tan\theta) = 2\tan\theta p+1p=(sec⁡θ+tan⁡θ)+(sec⁡θ−tan⁡θ)=2sec⁡θp + \frac{1}{p} = (\sec\theta + \tan\theta) + (\sec\theta - \tan\theta) = 2\sec\theta

Compute: p2−1p2+1=p−1pp+1p(dividing numerator and denominator by p)\frac{p^2 - 1}{p^2 + 1} = \frac{p - \frac{1}{p}}{p + \frac{1}{p}} \quad (\text{dividing numerator and denominator by } p) =2tan⁡θ2sec⁡θ=tan⁡θsec⁡θ=sin⁡θcos⁡θ1cos⁡θ=sin⁡θ= \frac{2\tan\theta}{2\sec\theta} = \frac{\tan\theta}{\sec\theta} = \frac{\frac{\sin\theta}{\cos\theta}}{\frac{1}{\cos\theta}} = \sin\theta

Hence proved.

Common mistake:
Not realizing sec⁡θ−tan⁡θ=1p\sec\theta - \tan\theta = \frac{1}{p}, so being unable to simplify the expression.
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How to approach Exercise 8.3

  1. Re-read the chapter summary first. Open Introduction to Trigonometry and refresh the key concepts: Trigonometric ratios, Standard angles, Pythagorean identity, Reciprocal identities.
  2. Try each problem yourself before opening the solution. Ten focused minutes per problem usually beats reading two finished solutions.
  3. Use Guru AI for the questions you get stuck on. The Socratic AI tutor walks you through it with hints instead of dictating the answer.
  4. Mark the questions you got wrong and revisit them after 24 hours — the spacing is what locks the method into long-term memory.

All exercises in Introduction to Trigonometry

  1. Exercise 8.1 — Trigonometric ratios for an acute angle in a right triangle.
  2. Exercise 8.2 — Trigonometric ratios at standard angles 0°, 30°, 45°, 60°, 90°.
  3. Exercise 8.3 — Trigonometric identities — sin²θ + cos²θ = 1 and consequences.

← Exercise 8.2

Trigonometric ratios at standard angles 0°, 30°, 45°, 60°, 90°.

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