Inverse Trigonometric Functions — Class 12 Maths Solution

Let ${\cos ^{ - 1}}\frac{{\sqrt 3 }}{2} = x \Rightarrow \frac{{\sqrt 3 }}{2} = \cos x$

As we know that the range of the principal value branch of ${\cos ^{ - 1}}$ is $[0,\;\pi ]$
Then, $\cos x = \frac{{\sqrt 3 }}{2} = \cos \frac{\pi }{6},\;\;where\;\;\frac{\pi }{6} \in [0,\;\pi ]$

Hence, the principal value of ${\cos ^{ - 1}}\frac{{\sqrt 3 }}{2}$ is $\frac{\pi }{6}.$