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Class 12 Inverse Trigonometric Functions · Long Answer3 Marksinverse_trigonometric_equations

Class 12 Maths · Inverse Trigonometric Functions · Long Answer #96

For x>0, solve \(\tan[\sec^{-1}(1/x)]=\sin(\tan^{-1}2)\).

✅ Step-by-Step Solution

Final Answer: x=√5/3
  1. The real arcsec domain together with x>0 gives 0<x≤1. Let θ=arcsec(1/x)∈[0,π/2).
  2. Then cos θ=x and tan θ=√(1−x²)/x, while sin(arctan2)=2/√5.
  3. Solve √(1−x²)/x=2/√5. Squaring yields 5−5x²=4x², so x²=5/9. The positive root is x=√5/3.
  4. This value lies in (0,1] and gives tan θ=2/√5, so x=√5/3 is the unique solution.

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] The real arcsec domain together with x>0 gives 0<x≤1. Let θ=arcsec(1/x)∈[0,π/2).
  • [1 mark] Then cos θ=x and tan θ=√(1−x²)/x, while sin(arctan2)=2/√5.
  • [1 mark] Solve √(1−x²)/x=2/√5. Squaring yields 5−5x²=4x², so x²=5/9. The positive root is x=√5/3.
  • [0.5 marks] This value lies in (0,1] and gives tan θ=2/√5, so x=√5/3 is the unique solution.

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