Question

If ${\cos }^{-1}x=\alpha ,(0<x<1)$ and ${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)+{\sec }^{-1}\left(\frac{1}{2x^2-1}\right)=\frac{2\pi }{3}$, then ${\tan }^{-1}(2x)$ equals

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (C)

${\cos }^{-1}x=\alpha \Rightarrow x=\cos \alpha$
${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)+{\sec }^{-1}\left(\frac{1}{2x^2-1}\right)=\frac{2\pi }{3}$
$2\alpha +2\alpha =\frac{2\pi }{3}$
$\Rightarrow 4\alpha =\frac{2\pi }{3}$
$\Rightarrow \alpha =\frac{\pi }{6}$
$x=\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$
${\tan }^{-1}(2x)={\tan }^{-1}(\sqrt{3})=\pi /3$