Question

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

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

Answer 1

Answer: (B)

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