Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Inverse Trigonometric Functions

Question:

The  value of $sin^{-1} \left[cos \begin{Bmatrix}sin^{-1} \left(-\frac{\sqrt{3}}{2}\right)\end{Bmatrix}\right]$, is

Options:

$\frac{\pi}{3}$

$\frac{\pi}{6}$

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

$-\frac{\pi}{6}$

Correct Answer:

$\frac{\pi}{6}$

Explanation:

$sin^{-1} \left[cos \begin{Bmatrix}sin^{-1} \left(-\frac{\sqrt{3}}{2}\right)\end{Bmatrix}\right]$

$sin^{-1}\begin{Bmatrix}cos \left(-\frac{\pi}{3}\right)\end {Bmatrix}$         $\left[∵sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)=-\frac{\pi}{3}\right]$

$= sin^{-1}\left(cos\frac{\pi}{3}\right) - sin^{-1} (\frac{1}{2}) =\frac{\pi}{6}$