Find the principal value of $cot^{-1} \left( -\dfrac{1}{\sqrt{3}} \right)$

The correct answer is $\dfrac{2\pi}{3}$

$cot^{-1} \left( -\dfrac{1}{\sqrt{3}} \right)$

= $π - cot^{-1} \left( \dfrac{1}{\sqrt{3}} \right)$

= $π - cot^{-1} \left( cot \dfrac{π}{3} \right)$

= $π - \dfrac{π}{3}$

= $\dfrac{2 π}{3}$