If $\sin ^{-1}\left(\frac{x}{5}\right)+\operatorname{cosec}^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2}$,then the value of $x$ is

If $\sin ^{-1} x - \cos ^{-1} x = \frac{\pi}{6}$, then $x$ is equal to

If $y = \sin^{-1}\left(\frac{19}{20}x\right) + \cos^{-1}\left(\frac{19}{20}x\right)$,then $\frac{dy}{dx} = $

Consider the following statements:
Assertion $(A)$: For $x \in \mathbb{R}-\{1\}$, $\frac{d}{dx}\left(\tan^{-1}\left(\frac{1+x}{1-x}\right)\right) = \frac{d}{dx}\left(\tan^{-1} x\right)$.
Reason $(R)$: For $x < 1$, $\tan^{-1}\left(\frac{1+x}{1-x}\right) = \frac{\pi}{4} + \tan^{-1} x$, and for $x > 1$, $\tan^{-1}\left(\frac{1+x}{1-x}\right) = -\frac{3\pi}{4} + \tan^{-1} x$.
The correct answer is:

Prove $\tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x$,where $-\frac{1}{\sqrt{2}} \leq x \leq 1$.

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