${\tan ^{ - 1}}\sqrt 3 - {\sec ^{ - 1}}( - 2)$ is equal to

(A) $\pi$

(B) $- \frac{\pi }{3}$

(C) $\frac{\pi }{3}$

(D) $\frac{{2\pi }}{3}$

Option B is correct

As we know that the range of principal value branch of ${\tan ^{ - 1}}$ and ${\sec ^{ - 1}}$ are $\left( { - \frac{\pi }{2},\;\frac{\pi }{2}} \right)$ and
$[0,\;\pi ] - \left\{ {\frac{\pi }{2}} \right\}$ respectively.

Let ${\tan ^{ - 1}}(\sqrt 3 ) = x \Rightarrow \sqrt 3 = \tan x,\;then,\;\sqrt 3 = \tan \left( {\frac{\pi }{3}} \right),\;where$ $\frac{\pi }{3} \in \left( { - \frac{\pi }{2},\;\frac{\pi }{2}} \right)$

Let ${\sec ^{ - 1}}( - 2) = y \Rightarrow - 2 = \sec y$
Then, $- 2 = - \sec \left( {\frac{\pi }{3}} \right) = \sec \left( {\pi - \frac{\pi }{3}} \right) = \sec \frac{{2\pi }}{3},$

Where $\frac{{2\pi }}{3} \in [0,\;\pi ] - \left\{ {\frac{\pi }{2}} \right\}$
$\therefore$ ${\tan ^{ - 1}}\sqrt 3 - {\sec ^{ - 1}}( - 2) = \frac{\pi }{3} - \frac{{2\pi }}{3} = \frac{\pi }{3}.$