Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Determinants

Question:

The value of $\tan^{-1}\left[2 \sin \{2 \cos^{-1} \left(\frac{\sqrt{3}}{2} \right) \}\right]$

Options:

$\frac{\pi}{3}$

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

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

$\frac{\pi}{6}$

Correct Answer:

$\frac{\pi}{3}$

Explanation:

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

$=\tan^{-1} \left[2 \sin \frac{2 × \pi}{6} \right]$

$=\tan^{-1} \left[2 \sin \frac{\pi}{3} \right]$

$= \tan^{-1}\left[2 × \frac{\sqrt{3}}{2}\right]$

$= \tan^{-1}[\sqrt{3}]$

$=\frac{\pi}{3}$