Let $x \in (0, 1)$. The set of all $x$ such that $\sin^{-1} x > \cos^{-1} x$ is the interval

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

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 $2 \sin ^{-1} \frac{3}{5} = \tan ^{-1} \frac{24}{7}$.

Let $M$ and $m$ respectively be the maximum and minimum values of the function $f(x) = \tan^{-1}(\sin x + \cos x)$ in the interval $[0, \frac{\pi}{2}]$. Then the value of $\tan(M - m)$ is equal to:

The value of $\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{8}\right)$ is