Solve: $\;$ $\tan^{-1} \left(\dfrac{2 x}{1 - x^2}\right) + \cot^{-1} \left(\dfrac{1 - x^2}{2x}\right) = \dfrac{\pi}{3}, \;$ where $\;$ $x > 0$
$\begin{aligned}
\tan^{-1} \left(\dfrac{2 x}{1 - x^2}\right) + \cot^{-1} \left(\dfrac{1 - x^2}{2x}\right) & = \tan^{-1} \left(\dfrac{2 x}{1 - x^2}\right) + \cot^{-1} \left(\dfrac{1}{\dfrac{2x}{1 - x^2}}\right) \\\\
& \left[\text{Note: } \cot^{-1} \left(\dfrac{1}{x}\right) = \tan^{-1} \left(x\right)\right] \\\\
& = \tan^{-1} \left(\dfrac{2 x}{1 - x^2}\right) + \tan^{-1} \left(\dfrac{2x}{1 - x^2}\right) \\\\
& = 2 \tan^{-1} \left(\dfrac{2x}{1 - x^2}\right)
\end{aligned}$
$\therefore \;$ $\tan^{-1} \left(\dfrac{2 x}{1 - x^2}\right) + \cot^{-1} \left(\dfrac{1 - x^2}{2x}\right) = \dfrac{\pi}{3}$
$\implies$ $2 \tan^{-1} \left(\dfrac{2x}{1 - x^2}\right) = \dfrac{\pi}{3}$
i.e. $\;$ $\tan^{-1} \left(\dfrac{2x}{1 - x^2}\right) = \dfrac{\pi}{6}$
i.e. $\;$ $\dfrac{2x}{1 - x^2} = \tan \left(\dfrac{\pi}{6}\right) = \dfrac{1}{\sqrt{3}}$
i.e. $\;$ $2 \sqrt{3} x = 1 - x^2$
i.e. $\;$ $x^2 + 2 \sqrt{3} x - 1 = 0$
i.e. $\;$ $x = \dfrac{- 2 \sqrt{3} \pm \sqrt{12 + 4}}{2}$
i.e. $\;$ $x = \dfrac{-2 \sqrt{3} \pm 4}{2}$
i.e. $\;$ $x = - \sqrt{3} + 2$ $\;$ OR $\;$ $x = - \sqrt{3} - 2$
Given: $\;$ $x > 0$
$\therefore \;$ $x = 2 - \sqrt{3}$