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

If $\tan ^{-1}\left(\frac{1}{3}\right) + \tan ^{-1}\left(\frac{1}{7}\right) + \tan ^{-1}\left(\frac{1}{13}\right) + \tan ^{-1}\left(\frac{1}{21}\right) + \tan ^{-1}\left(\frac{1}{31}\right) = \tan ^{-1}\left(\frac{p}{q}\right)$,where $p$ and $q$ are relatively prime numbers,then $p + q$ is equal to-

If $\cos^{-1} x + \cos^{-1} y + \cos^{-1} z = \pi$,then

The value of $x$,for which $\sin \left(\cot ^{-1}(x)\right)=\cos \left(\tan ^{-1}(1+x)\right)$,is

$2 \tan ^{-1}\left(\frac{1}{3}\right)-\tan ^{-1}\left(\frac{3}{4}\right)=$

Let the maximum value of $(\sin^{-1}x)^{2} + (\cos^{-1}x)^{2}$ for $x \in [-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}]$ be $\frac{m}{n}\pi^{2}$, where $\gcd(m, n) = 1$. Then $m+n$ is equal to ........... .