Inverse Trigonometric Functions — Class 12 Maths Solution

Let ${\sec ^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right) = x \Rightarrow \sec x = \frac{2}{{\sqrt 3 }}$

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

Then, $\frac{2}{{\sqrt 3 }} = \sec \left( {\frac{\pi }{6}} \right),\;\;where\;\frac{\pi }{6} \in [0,\;\pi ] - \left\{ {\frac{\pi }{2}} \right\}$

Hence, the principal value of ${\sec ^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right)\;\,is\;\;\frac{\pi }{6}.$