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

The value of $2 \tan ^{-1} \frac{1}{2} + \tan ^{-1} \frac{3}{8}$ is

$\sin ^{-1} \frac{\sqrt{3}}{2} + \sin ^{-1} \sqrt{\frac{2}{3}} = $

If $2\tan^{-1}(\cos x) = \tan^{-1}(2\csc x)$,then $x =$

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

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: