If $\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)=\frac{1}{2} \cos ^{-1} x$,then $x$ is

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 that $2 \tan ^{-1} \frac{1}{2} + \tan ^{-1} \frac{1}{7} = \tan ^{-1} \frac{31}{17}$.

The value of ${\tan ^{ - 1}}\left[ {\cos \left( {2\,{{\tan }^{ - 1}}\frac{3}{4}} \right)\, + \,\sin \,\left( {2\,{{\cot }^{ - 1}}\frac{1}{2}} \right)} \right]$ is

The sum of the first $n$ terms of the series $\cot^{-1} 3 + \cot^{-1} 7 + \cot^{-1} 13 + \cot^{-1} 21 + \dots$ is given by:

If ${x_1}, {x_2}, {x_3}, {x_4}$ are roots of the equation ${x^4} - {x^3}\sin 2\beta + {x^2}\cos 2\beta - x\cos \beta - \sin \beta = 0$,then ${\tan ^{ - 1}}{x_1} + {\tan ^{ - 1}}{x_2} + {\tan ^{ - 1}}{x_3} + {\tan ^{ - 1}}{x_4} = $