Prove that: $\;$ $\tan^{-1} \left(\dfrac{1}{7}\right) + \tan^{-1} \left(\dfrac{1}{13}\right) = \tan^{-1} \left(\dfrac{2}{9}\right)$
$\tan^{-1} \left(x\right) + \tan^{-1} \left(y\right) = \tan^{-1} \left(\dfrac{x + y}{1 - x \cdot y}\right), \;\; x \cdot y < 1$
$\begin{aligned}
\therefore \; \tan^{-1} \left(\dfrac{1}{7}\right) + \tan^{-1} \left(\dfrac{1}{13}\right) & = \tan^{-1} \left(\dfrac{\dfrac{1}{7} + \dfrac{1}{13}}{1 - \dfrac{1}{7} \times \dfrac{1}{13}}\right) \\\\
& = \tan^{-1} \left(\dfrac{20}{90}\right) \\\\
& = \tan^{-1} \left(\dfrac{2}{9}\right)
\end{aligned}$
Hence proved.