$A$ solution of the equation $\tan^{-1}(1 + x) + \tan^{-1}(1 - x) = \frac{\pi}{2}$ is

If $(\tan ^{-1} x)^2+(\cot ^{-1} x)^2=\frac{5 \pi^2}{8}$,then $x^2+1=$

$\sin \left[3 \sin ^{-1}(0.4)\right] = \ldots$

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-

The value of the expression $ \tan \left(\frac{1}{2} \cos ^{-1} \frac{2}{\sqrt{5}}\right) $ is

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