If $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1,$ then find the value of $x$.

The value of $\tan \left[2 \tan ^{-1}\left(\frac{1}{5}\right)-\frac{\pi}{4}\right]$ is

Prove $3 \sin ^{-1} x = \sin ^{-1}(3 x - 4 x^{3})$,where $x \in [-\frac{1}{2}, \frac{1}{2}]$.

If $\alpha = \cos^{-1}\left(\frac{3}{5}\right)$ and $\beta = \tan^{-1}\left(\frac{1}{3}\right)$,where $0 < \alpha, \beta < \frac{\pi}{2}$,then $\alpha - \beta$ is equal to

If $A=2 \tan ^{-1}\left(\frac{1+x}{1-x}\right)$ and $B=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)$,where $x \in(0,1)$,then $A-B=$

$A$ possible value of $\tan \left(\frac{1}{4} \sin ^{-1} \frac{\sqrt{63}}{8}\right)$ is :