Inverse Trigonometric Functions — Class 12 Maths Solution

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

As we know that the range of principal value of ${\cot ^{ - 1}}\;is\;(0,\;\pi )$

Then, $\sqrt 3 = \cot \left( {\frac{\pi }{6}} \right),\;\;where\;\frac{\pi }{6} \in (0,\;\pi )$

Hence, the principal value branch of ${\cot ^{ - 1}}(\sqrt 3 )\;\;is\;\frac{\pi }{6}.$