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

If $\frac{1}{2} \sin^{-1}\left(\frac{3 \sin 2\theta}{5+4 \cos 2\theta}\right) = \tan^{-1} x$,then $x =$

If $\cos ^{-1} x - \cos ^{-1} \frac{y}{3} = \alpha$,where $-1 \leq x \leq 1$,$-3 \leq y \leq 3$,and $x \leq \frac{y}{3}$,then for all $x, y$,$9x^2 - 6xy \cos \alpha + y^2$ is equal to

Evaluate: $\tan^{-1} \left( \frac{1}{\sqrt{x^2 - 1}} \right)$

If $0 \leqslant \cos ^{-1} x \leqslant \pi$ and $\frac{-\pi}{2} \leqslant \sin ^{-1} x \leqslant \frac{\pi}{2}$,then at $x=\frac{1}{5}$ the value of $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$ is

If $\sec ^{-1} \frac{x}{a}-\sec ^{-1} \frac{x}{b}=\sec ^{-1} b-\sec ^{-1} a$,then $x$ is equal to