Prove that $\;$ $2 \tan^{-1} \left(\dfrac{1}{3}\right) = \tan^{-1} \left(\dfrac{3}{4}\right)$
$\begin{aligned} \text{LHS } & = 2 \tan^{-1} \left(\dfrac{1}{3}\right) \\\\ & = \tan^{-1} \left(\dfrac{1}{3}\right) + \tan^{-1} \left(\dfrac{1}{3}\right) \\\\ & = \tan^{-1} \left(\dfrac{\dfrac{1}{3} + \dfrac{1}{3}}{1 - \dfrac{1}{3} \times \dfrac{1}{3}}\right) \\\\ & \left[\text{Note: } \tan^{-1} \left(x\right) + \tan^{-1} \left(y\right) = \tan^{-1} \left(\dfrac{x + y}{1 - x \cdot y}\right), \;\; x \cdot y < 1\right] \\\\ & = \tan^{-1} \left(\dfrac{\dfrac{2}{3}}{\dfrac{8}{9}}\right) \\\\ & = \tan^{-1} \left(\dfrac{3}{4}\right) = \text{ RHS} \end{aligned}$