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

If $\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}$,then the value of $x$ is

If $0 < x < 1$,then $\cot ^{-1}\left( \frac{2x^2 - 1}{2x\sqrt{1 - x^2}} \right)$ is equal to

${\sin ^{ - 1}}\left[ {x\sqrt {1 - x} - \sqrt x \sqrt {1 - {x^2}} } \right] = $

If $y = \sin^{-1} \left( \frac{\sqrt{1+x} + \sqrt{1-x}}{2} \right)$,then $\frac{dy}{dx} = $

$\tan \left[ {\frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{{2a}}{{1 + {a^2}}}} \right) + \frac{1}{2}{{\cos }^{ - 1}}\left( {\frac{{1 - {a^2}}}{{1 + {a^2}}}} \right)} \right] = $