If $\sin^{-1}\left(\tan\dfrac{\pi}{4}\right) - \sin^{-1}\left(\sqrt{\dfrac{3}{x}}\right) = \dfrac{\pi}{6}$ then $x$ is a root of the equation

For $x > 0$, if $\sin(\cos^{-1}x + \tan^{-1}x) - \cos(\sin^{-1}x + \tan^{-1}x) = \sin(\cot^{-1}2)$ then $x = $

If $\tan^{-1}ax + \tan^{-1}3x = \dfrac{\pi}{4}$, where $3ax^2 < 1$, then value of $a$ for $x = \dfrac{1}{6}$ is $\ldots$

$\sec^2(\tan^{-1}3) - \tan^2(\sec^{-1}3) = $