Question

Considering only the principal values of the inverse trigonometric functions, the value of
$\frac{3}{2} \cos ^{-1} \sqrt{\frac{2}{2+\pi^2}}+\frac{1}{4} \sin ^{-1} \frac{2 \sqrt{2} \pi}{2+\pi^2}+\tan ^{-1} \frac{\sqrt{2}}{\pi}$ is ___

Answer 1

$ \cos ^{-1} \sqrt{\frac{2}{2+\pi^2}}=\tan ^{-1} \frac{\pi}{\sqrt{2}}$

$ \sin ^{-1}\left(\frac{2 \sqrt{2} \pi}{2+\pi^2}\right)=\sin ^{-1}\left(\frac{2 \times \frac{\pi}{\sqrt{2}}}{1+\left(\frac{\pi}{\sqrt{2}}\right)^2}\right) $
$ =\pi-2 \tan ^{-1}\left(\frac{\pi}{\sqrt{2}}\right)$
$\left(\right.$ As, $\left.\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)=\pi-2 \tan ^{-1} x, x \geq 1\right)$
$ \text { and } \tan ^{-1} \frac{\sqrt{2}}{\pi}=\cot ^{-1}\left(\frac{\pi}{\sqrt{2}}\right)$
$ \therefore \text { Expression }=\frac{3}{2}\left(\tan ^{-1} \frac{\pi}{\sqrt{2}}\right)+\frac{1}{4}\left(\pi-2 \tan ^{-1} \frac{\pi}{\sqrt{2}}\right)+\cot ^{-1}\left(\frac{\pi}{\sqrt{2}}\right) $
$ =\left(\frac{3}{2}-\frac{2}{4}\right) \tan ^{-1} \frac{\pi}{\sqrt{2}}+\frac{\pi}{4}+\cot ^{-1} \frac{\pi}{\sqrt{2}} $
$=\left(\tan ^{-1} \frac{\pi}{\sqrt{2}}+\cot ^{-1} \frac{\pi}{\sqrt{2}}\right)+\frac{\pi}{4} $
$ =\frac{\pi}{2}+\frac{\pi}{4}=\frac{3 \pi}{4} $
$=2.35 \text { or } 2.36$