Target Exam

CUET

Subject

Section B1

Chapter

Inverse Trigonometric Functions

Question:

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

Options:

$-\frac{\pi}{3}$

$\frac{\pi}{3}$

$\frac{2\pi}{3}$

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

Correct Answer:

$\frac{2\pi}{3}$

Explanation:

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 $\frac{2\pi}{3}$.