If $\sin \left(\cot ^{-1}(x+1)\right)=\cos \left(\tan ^{-1} x\right)$,then the value of $x$ is equal to

Considering only the principal values of the inverse trigonometric functions,the value of $\frac{3}{2} \cos ^{-1} \sqrt{\frac{2}{2+\pi^2}}+\frac{1}{4} \sin ^{-1} \frac{2 \sqrt{2} \pi}{2+\pi^2}+\tan ^{-1} \frac{\sqrt{2}}{\pi}$ is:

Let $x \in (0, 1)$. The set of all $x$ such that $\sin^{-1} x > \cos^{-1} x$ is the interval

$S = \tan^{-1}\left( \frac{1}{n^2 + n + 1} \right) + \tan^{-1}\left( \frac{1}{n^2 + 3n + 3} \right) + \dots + \tan^{-1}\left( \frac{1}{1 + (n + 19)(n + 20)} \right)$,then $\tan S$ is equal to

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

The number of solutions of $\tan^{-1}4x + \tan^{-1}6x = \frac{\pi}{6}$ where $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$ is equal to