Solve: $\;$ $\tan^{-1} \left(2 x\right) + \tan^{-1} \left(\dfrac{1}{x + 4}\right) = \dfrac{\pi}{2}$
$\tan^{-1} x + \tan^{-1} y = \dfrac{\pi}{2} \iff x y = 1$
$\therefore \;$ $\tan^{-1} \left(2 x\right) + \tan^{-1} \left(\dfrac{1}{x + 4}\right) = \dfrac{\pi}{2}$ $\iff$ $2 x \times \dfrac{1}{x + 4} = 1$
i.e. $\;$ $2 x = x + 4$ $\implies$ $x = 4$
Verification:
Putting $\;$ $x = 4$ $\;$ in the given equation,
$LHS = \tan^{-1} \left(8\right) + \tan^{-1} \left(\dfrac{1}{4 + 4}\right)$
$= \tan^{-1} \left(8\right) + \tan^{-1} \left(\dfrac{1}{8}\right) = \dfrac{\pi}{2} = RHS$ $\;\;\;$ $\because \; 8 \times \dfrac{1}{8} = 1$
$\therefore \;$ The solution set is $\left\{4\right\}$