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| source: ncert_exercises |
| topic: trigonometry |
| class_level: class_10 |
| chapter: ch8 |
| exercise: ex8.4 |
| difficulty: advanced |
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| Class 10 | Ch8: Introduction to Trigonometry | Exercise 8.4 |
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| Proving Trigonometric Identities using: sin²θ+cos²θ=1, 1+tan²θ=sec²θ, 1+cot²θ=cosec²θ |
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| 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. ✅ |
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| 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. ✅ |
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| 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. ✅ |
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| 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. ✅ |
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| 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. ✅ |
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| 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. ✅ |
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| 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. ✅ |
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| 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. ✅ |
