If $\sec^{-1} x = \csc^{-1} y$,then $\cos^{-1} \frac{1}{x} + \cos^{-1} \frac{1}{y} = $

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:

The value of $\cot^{-1} \left[ \frac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}} \right]$, where $x \in (0, \frac{\pi}{2})$ is...

The differential coefficient of ${\tan ^{ - 1}}\left( \frac{{2x}}{{1 - {x^2}}} \right)$ with respect to ${\sin ^{ - 1}}\left( \frac{{2x}}{{1 + {x^2}}} \right)$ is

Evaluate: $\cos ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)$

If $\tan ^{-1}\left(\frac{1-x}{1+x}\right)-\frac{1}{2} \tan ^{-1} x=0$,for $x>0$,then $x=$