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Class 12 Inverse Trigonometric Functions · Long Answer3 Marksinverse_trigonometric_identities

Class 12 Maths · Inverse Trigonometric Functions · Long Answer #65

Prove \(\sin^{-1}(63/65)=\sin^{-1}(5/13)+\cos^{-1}(3/5)\).

✅ Step-by-Step Solution

Final Answer: Identity proved.
  1. Let α=arcsin(5/13), β=arccos(3/5). Both are acute, with cos α=12/13 and sin β=4/5.
  2. Then sin(α+β)=(5·3+12·4)/65=63/65.
  3. To fix the inverse branch, cos(α+β)=(12·3−5·4)/65=16/65>0. Since α+β∈(0,π), its positive cosine puts it in (0,π/2).
  4. Therefore α+β=arcsin(63/65), proving the identity.

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Let α=arcsin(5/13), β=arccos(3/5). Both are acute, with cos α=12/13 and sin β=4/5.
  • [1 mark] Then sin(α+β)=(5·3+12·4)/65=63/65.
  • [1 mark] To fix the inverse branch, cos(α+β)=(12·3−5·4)/65=16/65>0. Since α+β∈(0,π), its positive cosine puts it in (0,π/2).
  • [0.5 marks] Therefore α+β=arcsin(63/65), proving the identity.

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Prove \(\cos^{-1}(4/5)+\cos^{-1}(12/13)=\cos^{-1}(33/65)\).

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