Find the principal value of $\cot^{-1} \left( \frac{-1}{\sqrt{3}} \right)$. |
$-\frac{\pi}{3}$ $\frac{\pi}{3}$ $\frac{2\pi}{3}$ $\frac{5\pi}{6}$ |
$\frac{2\pi}{3}$ |
The correct answer is Option (3) → $\frac{2\pi}{3}$ ## Let $\cot^{-1} \left( \frac{-1}{\sqrt{3}} \right) = y$. Then, $\cot y = \frac{-1}{\sqrt{3}} = -\cot \left( \frac{\pi}{3} \right) = \cot \left( \pi - \frac{\pi}{3} \right) = \cot \left( \frac{2\pi}{3} \right)$ We know that the range of principal value branch of $\cot^{-1}$ is $(0, \pi)$ and $\cot \left( \frac{2\pi}{3} \right) = \frac{-1}{\sqrt{3}}$. Hence, principal value of $\cot^{-1} \left( \frac{-1}{\sqrt{3}} \right)$ is $\frac{2\pi}{3}$. |