Solve for $x$: $\tan^{-1}\left(\frac{1-x}{1+x}\right) = \frac{1}{2} \tan^{-1} x$,where $x > 0$.

If ${\sin ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - \dots} \right) + {\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - \dots} \right) = \frac{\pi }{2}$ for $0 < |x| < \sqrt 2$,then $x$ equals

Let the inverse trigonometric functions take principal values. The number of real solutions of the equation $2 \sin ^{-1} x + 3 \cos ^{-1} x = \frac{2 \pi}{5}$ is:

If $y=\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)+\tan ^{-1}\left(\frac{4 x-4 x^3}{1-6 x^2+x^4}\right)$,then $\frac{d y}{d x}$ is equal to

The value of $\tan \left(\sin ^{-1}\left(\frac{3}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right)$ is

The value of $\frac{\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)}{\operatorname{cosec}^{-1}(-\sqrt{2})+\cos ^{-1}\left(\frac{-1}{2}\right)}$ is