I have to simplify
$$\frac{3}{2}\cos^{-1}\left(\sqrt{\frac{2}{2+{\pi}^2}}\right)+\frac{1}{4} \sin^{-1} \left(\frac{2 \sqrt 2 \pi}{2+{\pi}^2}\right) +\tan^{-1}{\frac{\sqrt2}{\pi}}$$
My attempt:-
Let $\sqrt{\frac{2}{2+{\pi}^2}} =\cos{\theta}$
so that implies: $\sin{\theta}=\frac{\pi}{\sqrt{2+{\pi^2}}}$
similarly $\tan{\theta}=\frac{\pi}{\sqrt2}$
so the original expression simplifies to
$$\frac{3}{2} {\theta}+\frac{\theta}{2}+\left(\frac{\pi}{2}-{\theta}\right)$$
Which is $$ \theta +\frac{\pi}{2}$$
which gives $$\frac{\pi}{2} +\cos^{-1}\left(\frac{\sqrt{2}}{\sqrt{2+{\pi^2}}}\right)$$
however, the answer is supposed to be an integer, which this is clearly not.
Where am I going wrong?
Source:- JEE Advanced 2022, paper 1
Best Answer
The problem is due to this term $\sin^{-1} \left(\frac{2 \sqrt 2 \pi}{2+{\pi}^2}\right)$. Since the angle: $\frac{\pi}4<\theta<\frac{\pi}2$, so we have $\frac{\pi}2<2\theta<\pi$. But for inverse sine, its range is $\left[-\frac{\pi}2, \frac{\pi}2\right]$. Therefore,
$$\sin^{-1} \left(\frac{2 \sqrt 2 \pi}{2+{\pi}^2}\right)=\pi-2\theta$$