Question

If \[ \sin^{-1}(2x)-2\cos^{-1}\sqrt{1-x^2}=\frac{\pi}{2}, \] then \(\tan^{-1}(2x+1)=\)

• A. \(\dfrac{\pi}{6}\)
• B. \(\sin^{-1}x\)
• C. \(\dfrac{\pi}{3}\)
• D. \(\cos^{-1}x\)

Hint:

Useful identities: \[ \cos^{-1}\sqrt{1-x^2}=\sin^{-1}x, \] and \[ \tan^{-1}(\sqrt3)=\frac{\pi}{3}. \]

Answer 1

The Correct Option is C