Solve $2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \csc x)$

Let $S = \{ x : \cos^{-1} x = \pi + \sin^{-1} x + \sin^{-1}(2x + 1) \}$. Then $\sum_{x \in S} (2x - 1)^2$ is equal to . . . . . .

If $y = \cos \left(\frac{\pi}{3} + \cos^{-1} \frac{x}{2}\right)$,then $(x - y)^2 + 3y^2$ is equal to . . . . . . .

Let $\tan ^{-1}(x) \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,for $x \in R$. Then the number of real solutions of the equation $\sqrt{1+\cos (2 x)}=\sqrt{2} \tan ^{-1}(\tan x)$ in the set $\left(-\frac{3 \pi}{2},-\frac{\pi}{2}\right) \cup\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$ is equal to

$\tan \left(\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right) = $

$\sin ^{-1}\left(\frac{12}{13}\right)+\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{63}{16}\right)=$