QJEEAdv22Mab710: JEE Advanced 2022 Mathematics - Class 12 JEEAdv22Mab710 Solution: Considering only the principal values of the inverse…

Given, ${3 \over 2}{\cos ^{ - 1}}\sqrt {{2 \over {2 + {\pi ^2}}}} + {1 \over 4}{\sin ^{ - 1}}\left( {{{2\sqrt 2 \pi } \over {2 + {\pi ^2}}}} \right) + {\tan ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right)$

Let, ${\tan ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right) = \alpha$

$\Rightarrow {{\sqrt 2 } \over \pi } = \tan \alpha$

We know, $\sin 2\alpha = {{2\tan \alpha } \over {1 + {{\tan }^2}\alpha }}$

$= {{2\left( {{{\sqrt 2 } \over \pi }} \right)} \over {1 + {2 \over {{\pi ^2}}}}}$

$= {{2\sqrt 2 \pi } \over {2 + {\pi ^2}}}$

$\Rightarrow 2\alpha = {\sin ^{ - 1}}\left( {{{2\sqrt 2 \pi } \over {2 + {\pi ^2}}}} \right) = 2{\tan ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right)$

Now, let, ${\cos ^{ - 1}}\left( {\sqrt {{2 \over {2 + {\pi ^2}}}} } \right) = \beta$

$\Rightarrow \sqrt {{2 \over {2 + {\pi ^2}}}} = \cos \beta$

$\therefore$ $\cot \beta = {{\sqrt 2 } \over \pi }$

$\Rightarrow \beta = {\cot ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right)$

$\therefore$ ${\cos ^{ - 1}}\left( {\sqrt {{2 \over {2 + {\pi ^2}}}} } \right) = {\cot ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right)$

$\therefore$ ${3 \over 2}{\cos ^{ - 1}}\left( {\sqrt {{2 \over {2 + {\pi ^2}}}} } \right) + {1 \over 4}{\sin ^{ - 1}}\left( {{{2\sqrt 2 \pi } \over {2 + {\pi ^2}}}} \right) + {\tan ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right)$

$= {3 \over 2}{\cot ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right) + {1 \over 4}\left( {2 + \tan \left( {{{\sqrt 2 } \over \pi }} \right)} \right) + {\tan ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right)$

$= {3 \over 2}{\cot ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right) + {1 \over 2}{\tan ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right) + {\tan ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right)$

$= {3 \over 2}{\cot ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right) + {3 \over 2}{\tan ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right)$

$= {3 \over 2}\left( {{{\tan }^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right) + {{\cot }^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right)} \right)$

$= {3 \over 2} \times {\pi \over 2}$

$= {{3\pi } \over 4}$

$= 2.36$