$\cot \left(\sum_{n=1}^{50} \tan ^{-1}\left(\frac{1}{1+n+n^2}\right)\right) = $

Given $0 \leq x \leq \frac{1}{2}$,then the value of $\tan \left[\sin ^{-1}\left\{\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^{2}}}{\sqrt{2}}\right\}-\sin ^{-1} x\right]$ is:

$A$ possible value of $\tan \left(\frac{1}{4} \sin ^{-1} \frac{\sqrt{63}}{8}\right)$ is :

If $\sin^{-1}(x - 2) + \cos^{-1}(x) + \tan^{-1}(x + 2) + \cot^{-1}(x + 4) = \sec^{-1}(\sqrt{k}) - \frac{\pi}{2}$, then $\cos(2 \csc^{-1}\sqrt{k - 1}) = \dots$

If $\tan^{-1} (x+ 2)+ \tan^{- 1}( x -2)= \tan^{-1} (\frac{1}{2}),$ then the sum of the value$(s)$ of $x$ is equal to-

If $3 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)-4 \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)+2 \tan ^{-1}\left(\frac{2 x}{1-x^2}\right)=\frac{\pi}{3}$,then the value of $x=$