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 $\tan^{-1} \frac{1}{1+1(2)} + \tan^{-1} \frac{1}{1+2(3)} + \tan^{-1} \frac{1}{1+3(4)} + \dots + \tan^{-1} \frac{1}{1+n(n+1)} = \tan^{-1} \theta$,then $\theta$ =

Evaluate: $\operatorname{cosec}^{-1}\left[\left(\frac{\tan ^2\left(\frac{\alpha-\pi}{4}\right)-1}{\tan ^2\left(\frac{\alpha-\pi}{4}\right)+1}+\cos \frac{\alpha}{2} \cdot \cot 5 \alpha\right) \sec \frac{11 \alpha}{2}\right]$

$A$ value of $x$ satisfying the equation $\sin \left[ \cot^{-1} (1 + x) \right] = \cos \left[ \tan^{-1} x \right]$ is

If $\cos^{-1} x = \alpha$ $(0 < x < 1)$ and $\sin^{-1} (2 x \sqrt{1 - x^2}) + \sec^{-1} (\frac{1}{2 x^2 - 1}) = \frac{2 \pi}{3}$,then $\alpha$ is

If $\alpha$ and $\beta$ are the least and the greatest values of $f(x)=(\sin ^{-1} x)^2+(\cos ^{-1} x)^2$ for all $x \in [-1, 1]$ respectively,then $8(\alpha+\beta)=$