Question

The values of x satisfying $\tan \left({\sec }^{-1}x\right)=\sin \left({\cos }^{-1}\frac{1}{\sqrt{5}}\right)$ is

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

Answer 1

Answer: (B)

$\pm \frac{3}{\sqrt{5}}$

Let ${\cos }^{-1}\left(\frac{1}{\sqrt{5}}\right)=y$
$\cos \left(y\right)=\frac{1}{\sqrt{5}}$
Hence $\sin y=\frac{2}{\sqrt{5}}$
$\tan y=2$
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
${\sec }^2\left({\sec }^{-1}\right(x\left)\right)-1=\frac{4}{5}$
$x^2=\frac{9}{5}$
$\therefore x=\pm \frac{3}{\sqrt{5}}$