Question

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

• A. $\frac{5\pi}{6}$
• B. $\frac{\pi}{3}$
• C. $\frac{\pi}{2}$
• D. $\frac{\pi}{4}$

Answer 1

Answer: (A)

Let $cot^{-1} \left(-\sqrt{3}\right) = \theta$
$\Rightarrow cot\,\theta = -\sqrt{3}$
$=-cot \frac{\pi}{6}$
$= cot\left(\pi - \frac{\pi}{6}\right)$
$= cot \frac{5\pi}{6}$
$\Rightarrow \theta = \frac{5\pi }{6} \in \left(0,\,\pi\right) $
$\therefore $ Principal value of $cot^{-1} \left(-\sqrt{3}\right)$ is $\frac{5\pi }{6}$