Question

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

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

Answer 1

The correct option is B $\frac{\sqrt{5}}{3}$

$tan\left(se{c}^{-1}\frac{1}{x}\right)=sin\left(ta{n}^{-1}\right(2\left)\right)$
$tan\left(ta{n}^{-1}\right(\frac{\sqrt{1-x^2}}{x}\left)\right)=sin\left(si{n}^{-1}\right(\frac{2}{\sqrt{5}}\left)\right)$
$\frac{\sqrt{1-x^2}}{x}=\frac{2}{\sqrt{5}}$
$\frac{1-x^2}{x^2}=\frac{4}{5}$
$\frac{1}{x^2}-1=\frac{4}{5}$
$\frac{1}{x^2}=\frac{9}{5}$
$x^2=\frac{5}{9}$
$x=\pm \frac{\sqrt{5}}{3}$
Since $x>0$
$x=\frac{\sqrt{5}}{3}$