Question:

If \(\sin^{-1}(\frac{2a}{1+a^2})+\cos^{-1}(\frac{1-a^2}{1+a^2})=\tan^{-1}(\frac{2x}{1-x^2})\) where a, x ∈ (0, 1) then the value of x is

Updated On: Apr 17, 2024
  • \(\frac{2a}{1+a^2}\)
  • 0
  • \(\frac{2a}{1-a^2}\)
  • \(\frac{a}{2}\)
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The Correct Option is C

Solution and Explanation

The correct answer is (C) : \(\frac{2a}{1-a^2}\).
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