If $\sin ^{-1} x + \sin ^{-1} y = \frac{\pi}{2}$,then $x^{2}$ is equal to

The value of $\sin ^{-1}\left(-\frac{1}{\sqrt{2}}\right)+\cos ^{-1}\left(-\frac{1}{2}\right)-\cot ^{-1}\left(-\frac{1}{\sqrt{3}}\right)+\tan ^{-1}(\sqrt{3})$ is:

$2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{5 \sqrt{2}}{7}+2 \tan ^{-1} \frac{1}{8}=$

If $\operatorname{Tan}^{-1}\left[\frac{1}{1+1(2)}\right]+\operatorname{Tan}^{-1}\left[\frac{1}{1+(2)(3)}\right]+\operatorname{Tan}^{-1}\left[\frac{1}{1+(3)(4)}\right]+\cdots+\operatorname{Tan}^{-1}\left[\frac{1}{1+n(n+1)}\right]=\operatorname{Tan}^{-1} \theta$,then $\theta=$

The value of $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$ at $x=\frac{1}{5}$,where $0 \leq \cos ^{-1} x \leq \pi$ and $-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2}$,is

Evaluate: $\cot ^{ - 1}\left(\frac{xy + 1}{x - y}\right) + \cot ^{ - 1}\left(\frac{yz + 1}{y - z}\right) + \cot ^{ - 1}\left(\frac{zx + 1}{z - x}\right)$