Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

$\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 }$$=\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 }$

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

$=\frac{\pi }{2}+\frac{1}{2}{\tan }^{-1}\left(\frac{\pi }{\sqrt{2}}\right)-\frac{1}{4}{\tan }^{-1}\left(\frac{2·\left(\frac{\pi }{\sqrt{2}}\right)}{1-{\left(\frac{\pi }{\sqrt{2}}\right)}^2}\right)$

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

$=\frac{\pi }{2}+\frac{\pi }{4}=\frac{3\pi }{4}$

$=2.36$