Question:medium

If $3\sin^{-1}\!\left(\dfrac{2x}{1+x^{2}}\right) - 4\cos^{-1}\!\left(\dfrac{1-x^{2}}{1+x^{2}}\right) + 2\tan^{-1}\!\left(\dfrac{2x}{1-x^{2}}\right) = \dfrac{\pi}{3}$, then $x$ equals

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Key identities ($|x| \le 1$): $\sin^{-1}\tfrac{2x}{1+x^2} = 2\tan^{-1}x$; $\cos^{-1}\tfrac{1-x^2}{1+x^2} = 2\tan^{-1}x$; $\tan^{-1}\tfrac{2x}{1-x^2} = 2\tan^{-1}x$.
Updated On: Apr 8, 2026
  • $\dfrac{1}{\sqrt{3}}$
  • $-\dfrac{1}{\sqrt{3}}$
  • $\sqrt{3}$
  • $-\dfrac{\sqrt{3}}{2}$
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The Correct Option is A

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