Target Exam

CUET

Subject

Section B1

Chapter

Inverse Trigonometric Functions

Question:

The principal value of $\cos^{-1} \left( \frac{1}{2} \right) + \sin^{-1} \left( -\frac{1}{\sqrt{2}} \right)$ is:

Options:

$\frac{\pi}{12}$

$\pi$

$\frac{\pi}{3}$

$\frac{\pi}{6}$

Correct Answer:

$\frac{\pi}{12}$

Explanation:

The correct answer is Option (1) → $\frac{\pi}{12}$ ##

$\cos^{-1} \left( \frac{1}{2} \right) + \sin^{-1} \left( -\frac{1}{\sqrt{2}} \right)$

$= \cos^{-1} \left( \cos \frac{\pi}{3} \right) - \sin^{-1} \left( \frac{1}{\sqrt{2}} \right)$ [Since $\sin^{-1}(-\theta) = -\sin^{-1} \theta$]

$= \frac{\pi}{3} - \sin^{-1} \left( \sin \frac{\pi}{4} \right) = \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12}$