If $y=\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)+\tan ^{-1}\left(\frac{4 x-4 x^3}{1-6 x^2+x^4}\right)$, then $\frac{d y}{d x}$ is equal to

Prove $3 \cos ^{-1} x = \cos ^{-1} (4 x^{3} - 3 x)$,where $x \in [\frac{1}{2}, 1]$.

If $\theta = \tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{13}\right) + \tan^{-1}\left(\frac{1}{21}\right) + \tan^{-1}\left(\frac{1}{31}\right)$,then $\tan \theta =$

The derivative of $\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ with respect to $x$ is:

The value of $\tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)$ for $|x| < \frac{1}{\sqrt{2}}, x \neq 0$ is:

If $S = \{x \in R : \sin^{-1}\left(\frac{x+1}{\sqrt{x^2+2x+2}}\right) - \sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right) = \frac{\pi}{4}\}$,then $\sum_{x \in S} \left(\sin\left((x^2+x+5)\frac{\pi}{2}\right) - \cos((x^2+x+5)\pi)\right)$ is equal to $........$.