Prove that: $\;$ $\tan^{-1} \left(\dfrac{1}{5}\right) + \tan^{-1} \left(\dfrac{1}{3}\right) - \tan^{-1} \left(\dfrac{1}{7}\right) = \tan^{-1} \left(\dfrac{21}{53}\right)$
$\tan^{-1} \left(x\right) + \tan^{-1} \left(y\right) = \tan^{-1} \left(\dfrac{x + y}{1 - x \cdot y}\right), \;\; x \cdot y < 1$
$\tan^{-1} \left(x\right) - \tan^{-1} \left(y\right) = \tan^{-1} \left(\dfrac{x - y}{1 + x \cdot y}\right)$
$\begin{aligned}
\tan^{-1} \left(\dfrac{1}{5}\right) + \tan^{-1} \left(\dfrac{1}{3}\right) & = \tan^{-1} \left(\dfrac{\dfrac{1}{5} + \dfrac{1}{3}}{1 - \dfrac{1}{5} \times \dfrac{1}{3}}\right) \\\\
& = \tan^{-1} \left(\dfrac{8}{14}\right) \\\\
& = \tan^{-1} \left(\dfrac{4}{7}\right)
\end{aligned}$
$\begin{aligned}
\therefore \; \tan^{-1} \left(\dfrac{1}{5}\right) + \tan^{-1} \left(\dfrac{1}{3}\right) - \tan^{-1} \left(\dfrac{1}{7}\right) & = \tan^{-1} \left(\dfrac{4}{7}\right) - \tan^{-1} \left(\dfrac{1}{7}\right) \\\\
& = \tan^{-1} \left(\dfrac{\dfrac{4}{7} - \dfrac{1}{7}}{1 + \dfrac{4}{7} \times \dfrac{1}{7}}\right) \\\\
& = \tan^{-1} \left(\dfrac{21}{53}\right)
\end{aligned}$
Hence proved.