Compute the given expression: $\;$ $\tan^{-1} \left[- \tan \left(\dfrac{13 \pi}{8}\right)\right] + \cot^{-1} \left[\cot \left(\dfrac{-19 \pi}{6}\right)\right]$
Given expression: $\;$ $\tan^{-1} \left[- \tan \left(\dfrac{13 \pi}{8}\right)\right] + \cot^{-1} \left[\cot \left(\dfrac{-19 \pi}{6}\right)\right]$
$= - \tan^{-1} \left[\tan \left(\dfrac{13 \pi}{8}\right)\right] + \cot^{-1} \left[- \cot \left(\dfrac{19 \pi}{8}\right)\right]$
$\left\{\tan^{-1} \left(-x\right) = - \tan^{-1} x \;\; \forall \;\; x \in R\right\}$
$\left\{\cot \left(-x\right) = - \cot x\right\}$
$= - \tan^{-1} \left[\tan \left(\dfrac{13 \pi}{8}\right)\right] + \pi - \cot^{-1} \left[\cot \left(\dfrac{19 \pi}{8}\right)\right]$
$\left\{\cot^{-1} \left(-x\right) = \pi - \cot^{-1}x \;\; \forall \;\; x \in R\right\}$
$= - \tan^{-1} \left[\tan \left(2 \pi - \dfrac{3 \pi}{8}\right)\right] + \pi - \cot^{-1} \left[\cot \left(2 \pi + \dfrac{3 \pi}{8}\right)\right]$
$= - \tan^{-1} \left[- \tan \left(\dfrac{3 \pi}{8}\right)\right] + \pi - \cot^{-1} \left[\cot \left(\dfrac{3 \pi}{8}\right)\right]$
$\left\{\tan \left(2 \pi - \theta\right) = - \tan \theta, \;\;\; \cot \left(2 \pi + \theta\right) = \cot \theta\right\}$
$= - \left(-\right)\tan^{-1} \left[\tan \left(\dfrac{3 \pi}{8}\right)\right] + \pi - \cot^{-1} \left[\cot \left(\dfrac{3 \pi}{8}\right)\right]$
$= \tan^{-1} \left[\tan \left(\dfrac{3 \pi}{8}\right)\right] + \pi - \cot^{-1} \left[\cot \left(\dfrac{3 \pi}{8}\right)\right]$
$= \dfrac{3 \pi}{8} + \pi - \dfrac{3 \pi}{8}$
$\left\{\tan^{-1}\left(\tan x\right) = x \;\; \forall \;\; x \in \left(\dfrac{-\pi}{2}, \dfrac{\pi}{2}\right) \right\}$
$\left\{\cot^{-1} \left(\cot x\right) = x \;\; \forall \;\; x \in \left(0, \pi\right)\right\}$
$= \pi$