Question

The value of ${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)-{\sec }^{-1}\left(\frac{2}{\sqrt{3}}\right)$ is equal to

$\left(a\right)0\left(b\right)\frac{\pi }{6}$
$\left(c\right)\frac{\pi }{3}\left(d\right)\frac{\pi }{2}$

Answer 1

We know that, ${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6}$ and ${\sec }^{-1}\left(\frac{2}{\sqrt{3}}\right)=\frac{\pi }{6}$
$\therefore {\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)-{\sec }^{-1}\left(\frac{2}{\sqrt{3}}\right)=\left(\frac{\pi }{6}\right)-\left(\frac{\pi }{6}\right)=0$

Ans : $\left(a\right)$