If $\sin ^{ - 1}\frac{{2a}}{{1 + {a^2}}} - \cos ^{ - 1}\frac{{1 - {b^2}}}{{1 + {b^2}}} = \tan ^{ - 1}\frac{{2x}}{{1 - {x^2}}}$,then $x = $

$\tan^{-1}(-2) - \tan^{-1}(3)$ is equal to

If $\sin ^{-1}\left(\frac{x}{13}\right)+\operatorname{cosec}^{-1}\left(\frac{13}{12}\right)=\frac{\pi}{2},$ then the value of $x$ is

If $y = \tan^{-1} \left( \frac{4x}{1 + 5x^2} \right) + \tan^{-1} \left( \frac{2 + 3x}{3 - 2x} \right)$,then $\frac{dy}{dx} = $

$\frac{d}{dx} \left( \cos^{-1} \sqrt{\frac{1 + \cos x}{2}} \right) = $

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: