Prove that: $\;$ $2 \cot^{-1} \left(\dfrac{1}{3}\right) + \tan^{-1} \left(\dfrac{3}{4}\right) = \pi$
$\cot^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{1}{x}\right), \hspace{1cm} x > 0$
$\tan^{-1} \left(x\right) + \tan^{-1} \left(y\right) = \pi + \tan^{-1} \left(\dfrac{x + y}{1 - x \cdot y}\right), \;\; x \cdot y > 1$
$\tan^{-1} \left(-x\right) = - \tan^{-1} \left(x\right), \hspace{1cm} x \in R$
$\begin{aligned}
\therefore \; 2 \cot^{-1} \left(\dfrac{1}{3}\right) + \tan^{-1} \left(\dfrac{3}{4}\right) & = 2 \tan^{-1} \left(\dfrac{1}{1/3}\right) + \tan^{-1} \left(\dfrac{3}{4}\right) \\\\
& = 2 \tan^{-1} \left(3\right) + \tan^{-1} \left(\dfrac{3}{4}\right) \\\\
& \left[\because \; 3 \times 3 > 1\right] \\\\
& = \pi + \tan^{-1} \left(\dfrac{3 + 3}{1 - 3 \times 3}\right) + \tan^{-1} \left(\dfrac{3}{4}\right) \\\\
& = \pi + \tan^{-1} \left(- \dfrac{6}{8}\right) + \tan^{-1} \left(\dfrac{3}{4}\right) \\\\
& = \pi - \tan^{-1} \left(\dfrac{3}{4}\right) + \tan^{-1} \left(\dfrac{3}{4}\right) \\\\
& = \pi
\end{aligned}$
Hence proved.