The value of $\sin \left[\tan ^{-1}\left(\frac{1-x^2}{2 x}\right)+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]$ is

The trigonometric equation $\sin ^{-1} x = 2 \sin ^{-1} 2a$ has a real solution, if

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

If $\pi \le x \le 2\pi $,then ${\cos ^{ - 1}}(\cos x)$ is equal to

$\sin ^{-1} \frac{\sqrt{3}}{2} + \sin ^{-1} \sqrt{\frac{2}{3}} = $

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