$\cos ^{ - 1}\frac{4}{5} + \tan ^{ - 1}\frac{3}{5} = $

$\tan ^{-1} 2+\tan ^{-1} 3=$

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 $\alpha$ and $\beta$ are the least and the greatest values of $f(x)=(\sin ^{-1} x)^2+(\cos ^{-1} x)^2$ for all $x \in [-1, 1]$ respectively,then $8(\alpha+\beta)=$

If $\sin ^{-1}(4 x)+\sin ^{-1}(4 \sqrt{3} x)=-\frac{\pi}{2}$,then the absolute value of $x$ is

The sum to infinite terms of the series $\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{2}{9}\right) + \dots + \tan^{-1}\left(\frac{2^{n-1}}{1+2^{2n-1}}\right) + \dots$ is