The value of $\tan ^{-1} 2 + \tan ^{-1} 3$ is

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

Let $S$ be the set of all solutions of the equation $\cos ^{-1}(2 x)-2 \cos ^{-1}\left(\sqrt{1-x^2}\right)=\pi$,where $x \in\left[-\frac{1}{2}, \frac{1}{2}\right]$. Then $\sum_{x \in S} 2 \sin ^{-1}\left(x^2-1\right)$ is equal to

If $3 \sin^{-1}(\frac{2x}{1+x^2}) - 4 \cos^{-1}(\frac{1-x^2}{1+x^2}) + 2 \tan^{-1}(\frac{2x}{1-x^2}) = \frac{\pi}{3}$, then find the value of $x$.

$2 \pi - \left(\sin ^{-1} \frac{4}{5} + \sin ^{-1} \frac{5}{13} + \sin ^{-1} \frac{16}{65}\right)$ is equal to

If $\sin ^{-1} a=\alpha+\beta$ and $\sin ^{-1} b=\alpha-\beta$ then,$\sin ^2 \alpha+\cos ^2 \beta=$ . . . . . . .