$3 \tan^{-1} a$ is equal to

If $\alpha > \beta > \gamma > 0$,then the expression $\cot ^{-1}\left\{\beta+\frac{(1+\beta^2)}{(\alpha-\beta)}\right\}+\cot ^{-1}\left\{\gamma+\frac{(1+\gamma^2)}{(\beta-\gamma)}\right\}+\cot ^{-1}\left\{\alpha+\frac{(1+\alpha^2)}{(\gamma-\alpha)}\right\}$ is equal to:

$\cos ^{-1}\left(\frac{-1}{2}\right)-2 \sin ^{-1}\left(\frac{1}{2}\right)+3 \cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)-4 \tan ^{-1}(-1)$ equals

If $0 \le x \le 1$ and $(\sin^{-1} x)^3 + (\cos^{-1} x)^3 = a\pi^3$, then find the range of $a$.

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

$\sin^{-1}x + \sin^{-1}\frac{1}{x} + \cos^{-1}x + \cos^{-1}\frac{1}{x} = $