Check the given equality: $\;$ $\cot^{-1} \left(\dfrac{1}{7}\right) + 2 \cot^{-1} \left(\dfrac{1}{3}\right) = \dfrac{5 \pi}{4}$
$\begin{aligned} LHS & = \cot^{-1} \left(\dfrac{1}{7}\right) + 2 \cot^{-1} \left(\dfrac{1}{3}\right) \\\\ & = \tan^{-1} \left(7\right) + 2 \tan^{-1} \left(3\right) \;\;\; \left\{\because \;\; \tan^{-1} \left(\dfrac{1}{x}\right) = \cot^{-1}\left(x\right), \; x >0\right\} \\\\ & = \left[\tan^{-1} \left(7\right) + \tan^{-1} \left(3\right)\right] + \tan^{-1} \left(3\right) \\\\ & = \pi + \tan^{-1} \left(\dfrac{7 + 3}{1 - 7 \times 3}\right) + \tan^{-1} \left(3\right) \\\\ & \left\{\because \;\; \tan^{-1} x + \tan^{-1} y = \pi + \tan^{-1} \left(\dfrac{x + y}{1 - xy}\right), \; x > 0, \; y > 0, \; xy >1\right\} \\\\ & = \pi + \tan^{-1} \left(\dfrac{10}{-20}\right) + \tan^{-1} \left(3\right) \\\\ & = \pi + \tan^{-1} \left(\dfrac{-1}{2}\right) + \tan^{-1} \left(3\right) \\\\ & = \pi + \tan^{-1} \left(\dfrac{\dfrac{-1}{2} + 3}{1 - \left(\dfrac{-1}{2}\right) \times 3}\right) \\\\ & \left\{\because \;\; \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left(\dfrac{x + y}{1 - xy}\right), \;\; x y < 1\right\} \\\\ & = \pi + \tan^{-1} \left(\dfrac{\dfrac{5}{2}}{\dfrac{5}{2}}\right) \\\\ & = \pi + \tan^{-1} \left(1\right) \\\\ & = \pi + \dfrac{\pi}{4} \\\\ & = \dfrac{5 \pi}{4} = RHS \end{aligned}$