Target Exam

CUET

Subject

Maths. Section B1

Chapter

Inverse Trigonometric Functions

Question:

For the principal value branch, the value of $\sin\left(\frac{\pi}{2}-\sin^{-1}(-\frac{\sqrt{3}}{2})\right)$ is

Options:

$\frac{1}{\sqrt{2}}$

$\frac{\sqrt{3}}{2}$

$\frac{1}{2}$

$-\frac{\sqrt{3}}{2}$

Correct Answer:

$\frac{1}{2}$

Explanation:

The correct answer is Option (3) → $\frac{1}{2}$

Given expression: $\sin\left(\frac{\pi}{2}-\sin^{-1}(-\frac{\sqrt{3}}{2})\right)$

Use identity: $\sin\left(\frac{\pi}{2} - \theta\right) = \cos\theta$

$= \cos\left(\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right)$

Let $\theta = \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)$

Then, $\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - \left(-\frac{\sqrt{3}}{2}\right)^2} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$