If $\sin ^{-1} x < \cos ^{-1} x$, then

The value of ${\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right]$,where $|x| < 1$ and $x \ne 0$,is equal to

Assertion $(A): \operatorname{cosech}^{-1}(3) = \log \left(\frac{1+\sqrt{10}}{3}\right)$
Reason $(R): e^{\operatorname{cosech}^{-1} x}$ is a root of the quadratic equation $x p^2 - 2p - x = 0$
The correct option among the following is

Prove $\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2}$,where $x \in\left(0, \frac{\pi}{4}\right)$.

If $2 \sin ^{-1} x-3 \cos ^{-1} x=4, x \in[-1,1]$,then $2 \sin ^{-1} x+3 \cos ^{-1} x$ is equal to

If $\cos ^{-1}\left(\frac{12}{13}\right)+\sin ^{-1}\left(\frac{3}{5}\right)=\sin ^{-1} P$,then the value of $P$ is