If $\frac{(x+1)^{2}}{x^{3}+x}=\frac{A}{x}+\frac{Bx+C}{x^{2}+1}$,then $\sin^{-1} A + \tan^{-1} B + \sec^{-1} C$ is equal to

If $3{\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} - 4{\cos ^{ - 1}}\frac{{1 - {x^2}}}{{1 + {x^2}}} + 2{\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}} = \frac{\pi }{3}$,then $x$ =

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:

If $\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$,then the values of $x$ are

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 $x=\operatorname{cosec}(\tan ^{-1}(\cos (\cot ^{-1}(\sec (\sin ^{-1} a)))))$,where $a \in [0, 1]$,then which of the following is true?