Inverse Trigonometric Functions — Class 12 Maths Solution

Let ${\tan ^{ - 1}}( - \sqrt 3 ) = x \Rightarrow - \sqrt 3 = \tan x$

As we know that the range of the principal value branch of ${\tan ^{ - 1}}$ is $\left( { - \frac{\pi }{2},\;\frac{\pi }{2}} \right)$

Then, $\tan x = - \sqrt 3 = \tan \left( { - \frac{\pi }{3}} \right),\;\;where - \frac{\pi }{3} \in \left( { - \frac{\pi }{2},\;\frac{\pi }{2}} \right)$

Hence, the principal value of ${\tan ^{ - 1}}( - \sqrt 3 )$ is $- \frac{\pi }{3}.$