Question

if $\tan \left({\sec }^{-1}x\right)=\sin \left({\cos }^{-1}\frac{1}{\sqrt{5}}\right)$ then $x$=

• A. $\pm \frac{3}{\sqrt{5}}$
• B. $\pm \frac{\sqrt{5}}{3}$
• C. $\pm \frac{\sqrt{3}}{5}$
• D. none of these

Answer 1

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

Given, $\tan \left({\sec }^{-1}x\right)=\sin \left({\cos }^{-1}\frac{1}{\sqrt{5}}\right)$

${\cos }^{-1}\left(\frac{1}{\sqrt{5}}\right)=y$
$\cos \left(y\right)=\frac{1}{\sqrt{5}}$
Hence, $\sin y=\frac{2}{\sqrt{5}}$
$tany=2$
Hence, $\sin \left({\cos }^{-1}\right(\frac{1}{\sqrt{5}})$
$=\sin \left({\sin }^{-1}\right(\frac{2}{\sqrt{5}})$
$=\frac{2}{\sqrt{5}}$
Squaring on both the sides we get
${\sec }^2\left({\sec }^{-1}\right(x\left)\right)-1=\frac{4}{5}$
$x^2=\frac{9}{5}$
$\therefore x=\pm \frac{3}{\sqrt{5}}$