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