If $\sin ^{-1} x - \cos ^{-1} x = \frac{\pi}{6}$, then $x$ is equal to

$A$ value of $x$ satisfying the equation $\sin \left[ \cot^{-1} (1 + x) \right] = \cos \left[ \tan^{-1} x \right]$ is

$ \cos \left[2 \sin ^{-1} \frac{3}{4} + \cos ^{-1} \frac{3}{4}\right] $

If $x \geq 1$,then $2 \tan^{-1} x + \sin^{-1} (\frac{2x}{1+x^2})$ is equal to:

If for $x \in \left(0, \frac{1}{4}\right)$,the derivative of $\tan^{-1}\left(\frac{6x\sqrt{x}}{1-9x^3}\right)$ is $\sqrt{x} \cdot g(x)$,then $g(x)$ equals

The value of $\tan ^2(\sec ^{-1} 4) + \cot ^2(\operatorname{cosec}^{-1} 3)$ is