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)

$\frac{5\pi}{6}$

Answer 2

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 $\frac{5\pi }{6}$