Question

The value of $x$, where $x>0$ and $\tan \left({\sec }^{-1}\left(\frac{1}{x}\right)\right)=\sin \left({\tan }^{-1}2\right)$

• A. $\sqrt{5}$
• B. $\frac{\sqrt{5}}{3}$
• C. $1$
• D. 2 / 3

Answer 1

Answer: (B)

Given, $\tan \left({\sec }^{-1}\left(\frac{1}{x}\right)\right)=\sin \left({\tan }^{-1}2\right)$
$\Rightarrow \tan \left({\tan }^{-1}\frac{\sqrt{1-x^2}}{x}\right)$
$=\sin \left({\sin }^{-1}\frac{2}{\sqrt{1+2^2}}\right)$
$\Rightarrow \left(\because {\tan }^{-1}x={\sin }^{-1}\frac{x}{\sqrt{1+x^2}}\right)$
$\Rightarrow \frac{\sqrt{1-x^2}}{x}=\frac{2}{\sqrt{5}}$
$\Rightarrow \sqrt{5}\sqrt{1-x^2}=2x$
$\Rightarrow 4x^2=5\left(1-x^2\right)$
$\Rightarrow x^2=\frac{5}{9}$
$\Rightarrow x=\frac{\sqrt{5}}{3}$