$\operatorname{Tan}^{-1} \left( \frac{\sqrt{8-2 \sqrt{15}}}{\sqrt{15}+1} \right) + \operatorname{Tan}^{-1} \left( \frac{1}{\sqrt{5}} \right) =$

If $(\tan^{-1} x)^2 + (\cot^{-1} x)^2 = \frac{5\pi^2}{8}$,then $x$ =

If $\cos ^{-1} x - \cos ^{-1} \frac{y}{2} = \alpha$,where $-1 \leq x \leq 1, -2 \leq y \leq 2, x \leq \frac{y}{2}$,then for all $x, y$,$4x^2 - 4xy \cos \alpha + y^2$ is equal to

Consider the following statements:
Assertion $(A)$: When $x, y, z$ are positive numbers,then $\operatorname{Tan}^{-1}\left(\sqrt{\frac{x(x+y+z)}{y z}}\right)+\operatorname{Tan}^{-1}\left(\sqrt{\frac{y(x+y+z)}{x z}}\right)+\operatorname{Tan}^{-1}\left(\sqrt{\frac{z(x+y+z)}{x y}}\right) = \pi$
Reason $(R)$: $\operatorname{Tan}^{-1} a + \operatorname{Tan}^{-1} b = \operatorname{Tan}^{-1}\left(\frac{a+b}{1-ab}\right)$ if $a > 0$ and $b > 0$ and $ab < 1$.

$\cos \left(2 \cos ^{-1} \frac{1}{5}+\sin ^{-1} \frac{1}{5}\right)$ is equal to

Prove $\tan ^{-1}\left(\frac{2}{11}\right)+\tan ^{-1}\left(\frac{7}{24}\right)=\tan ^{-1}\left(\frac{1}{2}\right)$