Class 12 Inverse Trigonometric Functions · Long Answer3 Marksinverse_trigonometric_identities
Class 12 Maths · Inverse Trigonometric Functions · Long Answer #95
Prove \(2\sin^{-1}(3/5)-\tan^{-1}(17/31)=\pi/4\).
✅ Step-by-Step Solution
Final Answer: Identity proved.
- Put α=arcsin(3/5), β=arctan(17/31), both acute. Then tan α=3/4.
- Thus tan2α=2(3/4)/(1−9/16)=24/7. Also α<π/4, so 2α∈(0,π/2).
- Apply the tangent difference formula: tan(2α−β)=[24/7−17/31]/[1+(24/7)(17/31)]=(625/217)/(625/217)=1. Since 2α,β∈(0,π/2), their difference lies in (−π/2,π/2).
- Therefore 2α−β=π/4, proving the identity.
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] Put α=arcsin(3/5), β=arctan(17/31), both acute. Then tan α=3/4.
- [1 mark] Thus tan2α=2(3/4)/(1−9/16)=24/7. Also α<π/4, so 2α∈(0,π/2).
- [1 mark] Apply the tangent difference formula: tan(2α−β)=[24/7−17/31]/[1+(24/7)(17/31)]=(625/217)/(625/217)=1. Since 2α,β∈(0,π/2), their difference lies in (−π/2,π/2).
- [0.5 marks] Therefore 2α−β=π/4, proving the identity.
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