class 12 maths inverse trigonometric functions

${\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right)$

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📘 Inverse Trigonometric Functions NCERT Ex. 2.1, Q. 2 , Page 41 SA

${\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right)$

Official Solution

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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 $\frac{\pi }{6}.$

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