Solve: $\;$ $5 \tan^{-1} x + 3 \cot^{-1} x = 2 \pi$
Given equation: $\;$ $5 \tan^{-1} x + 3 \cot^{-1} x = 2 \pi$
i.e. $\;$ $2 \tan^{-1} x + \left(3 \tan^{-1} x + 3 \cot^{-1} x\right) = 2 \pi$
i.e. $\;$ $2 \tan^{-1} x + 3 \left(\tan^{-1} x + \cot^{-1} x \right) = 2 \pi$
i.e. $\;$ $2 \tan^{-1} x + 3 \times \dfrac{\pi}{2} = 2 \pi$ $\;\;\;$ $\left[\because \; \tan^{-1} x + \cot^{-1} x = \dfrac{\pi}{2}, \;\; \forall \; x \in R\right]$
i.e. $\;$ $2 \tan^{-1} x = \dfrac{\pi}{2}$
i.e. $\;$ $\tan^{-1} x = \dfrac{\pi}{4}$
i.e. $\;$ $x = \tan \left(\dfrac{\pi}{4}\right) = 1$