Question

The value of ${\tan }^{-1}\left(2\sin \left({\sec }^{-1}\left(2\right)\right)\right)$ is

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

Answer 1

The correct option is C $\frac{\pi }{3}$

${\tan }^{-1}\left(2\sin \left({\sec }^{-1}2\right)\right)$

Let ${\sec }^{-1}\left(2\right)=p$

$2=\sec p$

$\cos p=\frac{1}{2}$

$\Rightarrow p=\frac{\pi }{3}\Rightarrow {\sec }^{-1}\left(2\right)=\frac{\pi }{3}$

$\Rightarrow {\tan }^{-1}\left(2\sin \left(\frac{\pi }{3}\right)\right)(since{\sec }^{-1}\left(2\right)=\frac{\pi }{3})$

$\Rightarrow {\tan }^{-1}(2\times \frac{\sqrt{3}}{2})$

$\Rightarrow {\tan }^{-1}\left(\sqrt{3}\right)$

Let ${\tan }^{-1}\left(\sqrt{3}\right)=k$

$\Rightarrow \tan k=\sqrt{3}$

$\Rightarrow k=\frac{\pi }{3}$

$\therefore {\tan }^{-1}\left(\sqrt{3}\right)=\frac{\pi }{3}$

$\therefore {\tan }^{-1}\left(2\sin \left({\sec }^{-1}\left(2\right)\right)\right)=\frac{\pi }{3}$