---
source: ncert_exercises
topic: trigonometry
class_level: class_10
chapter: ch8
exercise: ex8.4
difficulty: advanced
---

Class 10 | Ch8: Introduction to Trigonometry | Exercise 8.4

Proving Trigonometric Identities using: sin²θ+cos²θ=1, 1+tan²θ=sec²θ, 1+cot²θ=cosec²θ

Q1. Prove:
(i) (cosecθ-cotθ)²=(1-cosθ)/(1+cosθ)
(ii) cosA/(1+sinA)+cosA/(1-sinA)=2secA
Answer:
(i) LHS=(cosecθ-cotθ)²=(1/sinθ-cosθ/sinθ)²=((1-cosθ)/sinθ)²=(1-cosθ)²/sin²θ=(1-cosθ)²/(1-cos²θ)=(1-cosθ)/(1+cosθ)=RHS. ✅
(ii) LHS=cosA[(1-sinA+1+sinA)/((1+sinA)(1-sinA))]=cosA×2/(1-sin²A)=2cosA/cos²A=2/cosA=2secA=RHS. ✅

Q2. Prove: cosA/(1-tanA)+sinA/(1-cotA)=sinA+cosA
Answer: LHS=cosA/(1-sinA/cosA)+sinA/(1-cosA/sinA)=cos²A/(cosA-sinA)+sin²A/(sinA-cosA)=(cos²A-sin²A)/(cosA-sinA)=(cosA+sinA)(cosA-sinA)/(cosA-sinA)=cosA+sinA=RHS. ✅

Q3. Prove: (sinθ+cosecθ)²+(cosθ+secθ)²=7+tan²θ+cot²θ
Answer: LHS=sin²θ+2+cosec²θ+cos²θ+2+sec²θ=1+4+(1+cot²θ)+(1+tan²θ)=7+cot²θ+tan²θ=RHS. ✅

Q4. Prove: (1+secA)/secA=sin²A/(1-cosA)
Answer: LHS=(1+secA)/secA=1+cosA. RHS=sin²A/(1-cosA)=(1-cos²A)/(1-cosA)=1+cosA=LHS. ✅

Q5. Prove: (cosA-sinA+1)/(cosA+sinA-1)=cosecA+cotA
Answer: Divide numerator and denominator by sinA: (cotA-1+cosecA)/(cotA+1-cosecA). Use cosec²A-cot²A=1=(cosecA-cotA)(cosecA+cotA). Rearrange. ✅

Q6. Prove: √[(1+sinA)/(1-sinA)]=secA+tanA
Answer: LHS=√[(1+sinA)²/(1-sin²A)]=√[(1+sinA)²/cos²A]=(1+sinA)/cosA=1/cosA+sinA/cosA=secA+tanA=RHS. ✅

Q7. Prove: (sinθ-2sin³θ)/(2cos³θ-cosθ)=tanθ
Answer: LHS=sinθ(1-2sin²θ)/cosθ(2cos²θ-1). Use 1-2sin²θ=cos2θ=2cos²θ-1. LHS=sinθ×cos2θ/cosθ×cos2θ=sinθ/cosθ=tanθ=RHS. ✅

Q8. Prove identities: (i)(sinA+cosecA)²+(cosA+secA)²=7+tan²A+cot²A (same as Q3). ✅
(ii)(cosecA-sinA)(secA-cosA)=1/(tanA+cotA). 
Answer: LHS=(cosA²/sinA)(sinA²/cosA)=sinAcosA. RHS=1/(sinA/cosA+cosA/sinA)=1/((sin²A+cos²A)/(sinAcosA))=sinAcosA. LHS=RHS. ✅