All the values of $x$ satisfying the equation $2 \tan^{-1} 2x = \sin^{-1} \left( \frac{4x}{1+4x^2} \right)$ lie in the interval

Statement $I:$ The equation $(\sin^{-1} x)^3 + (\cos^{-1} x)^3 - a\pi^3 = 0$ has a solution for all $a \ge \frac{1}{32}.$
Statement $II:$ For any $x \in [-1, 1],$ $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ and $0 \le (\sin^{-1} x - \frac{\pi}{4})^2 \le \frac{9\pi^2}{16}.$

If $\sin ^{-1} x < \cos ^{-1} x$, then

If $(\tan^{-1} x)^2 + (\cot^{-1} x)^2 = \frac{5\pi^2}{8}$, then the value of $x$ is equal to...

${\sin ^{ - 1}}\left[ {x\sqrt {1 - x} - \sqrt x \sqrt {1 - {x^2}} } \right] = $

If $\tan ^{-1}\left[\frac{1}{1+1 \cdot 2}\right]+\tan ^{-1}\left[\frac{1}{1+2 \cdot 3}\right]+\cdots+\tan ^{-1}\left[\frac{1}{1+n(n+1)}\right]=\tan ^{-1}[x]$,then $x=$