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} $
$\Rightarrow x=\frac{\sqrt{5}}{3}$