Solve the equation: tan⁻¹x - cot⁻¹x = tan⁻¹(1/(√3))

$\cot^{-1}x = \dfrac{\pi}{2} - \tan^{-1}x$, so $\tan^{-1}x - \cot^{-1}x = 2\tan^{-1}x - \dfrac{\pi}{2}$. $\tan^{-1}\left(\dfrac{1}{\sqrt{3}}\right) = \dfrac{\pi}{6}$, so $2\tan^{-1}x = \dfrac{\pi}{6} + \dfrac{\pi}{2} = \dfrac{2\pi}{3}$, $\tan^{-1}x = \dfrac{\pi}{3}$ and $x = \sqrt{3}$.