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

Answer: 2.36

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

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