JEE Advanced 2022 Paper 1, Mathematics Section 1 Q1: Inverse Trigonometric Functions

Let $\alpha ={\tan }^{-1}\frac{\pi }{\sqrt{2}}$, so $\cos \alpha =\sqrt{\frac{2}{2+{\pi }^2}}$, $\sin \alpha =\frac{\pi }{\sqrt{2+{\pi }^2}}$ and $\tan \alpha =\frac{\pi }{\sqrt{2}}$.

Using ${\cos }^{-1}\sqrt{\frac{2}{2+{\pi }^2}}=\alpha$ and ${\sin }^{-1}\frac{2\sqrt{2}\pi }{2+{\pi }^2}={\sin }^{-1}(2\sin \alpha \cos \alpha )$, the original sum becomes $\frac{3}{2}{\tan }^{-1}\frac{\pi }{\sqrt{2}}+\frac{1}{4}{\tan }^{-1}\left(\frac{2\sqrt{2}\pi }{{\pi }^2-2}\right)+{\tan }^{-1}\frac{\sqrt{2}}{\pi }$.

Since ${\tan }^{-1}\frac{\pi }{\sqrt{2}}+{\tan }^{-1}\frac{\sqrt{2}}{\pi }=\frac{\pi }{2}$, the expression simplifies to $\frac{\pi }{2}+\frac{1}{2}{\tan }^{-1}\frac{\pi }{\sqrt{2}}-\frac{1}{4}{\tan }^{-1}\left(\frac{2(\pi /\sqrt{2})}{1-(\pi /\sqrt{2}{)}^2}\right)$.

The double-angle identity gives ${\tan }^{-1}\left(\frac{2(\pi /\sqrt{2})}{1-(\pi /\sqrt{2}{)}^2}\right)=-\pi +2{\tan }^{-1}\frac{\pi }{\sqrt{2}}$ (as $\pi /\sqrt{2}>1$).

Substituting and simplifying, the answer becomes $\frac{\pi }{2}+\frac{\pi }{4}=\frac{3\pi }{4}\approx 2.36$.

$\therefore$ Required value $=2.36$.