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 ${\tan }^{-1}\left(2x\right)$ is equal to

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

Answer 1

Answer: (C)

$\frac{\pi }{3}$

Given that, ${\cos }^{-1}x=\alpha ,0<x<1$ .....(i)
$\Rightarrow x=\cos \alpha$
Thus ${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)+{\sec }^{-1}\left(\frac{1}{2x^2-1}\right)=\frac{2\pi }{3}$
$\Rightarrow {\sin }^{-1}\left(2\cos \alpha \sqrt{1-{\cos }^2\alpha }\right)+{\sec }^{-1}\left(\frac{1}{2{\cos }^2\alpha -1}\right)=\frac{2\pi }{3}$
$\Rightarrow {\sin }^{-1}\left(\sin 2\alpha \right)+{\sec }^{-1}\left(\sec 2\alpha \right)=\frac{2\pi }{3}$
$\Rightarrow 2\alpha +2\alpha =\frac{2\pi }{3}$
$\Rightarrow \alpha =\frac{\pi }{6}$ ....[from equation (i)]
Now, $x=\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$
$\Rightarrow 2x=\sqrt{3}$
Therefore, ${\tan }^{-1}\left(2x\right)={\tan }^{-1}\left(\sqrt{3}\right)=\frac{\pi }{3}$