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  • Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise  Very short answer question 43 math

Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise  Very short answer question 43 math

Answers (1)

Answer:  \frac{\pi}{3}

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

Hint: Try to solve \cos ^{-1}  and then \sin function.

Solution:

\begin{aligned} \tan ^{-1}\left[2 \sin \left(2 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right] &=\tan ^{-1}\left\{2 \sin \left[\cos ^{-1} 2\left(\frac{\sqrt{3}}{2}\right)^{2}-1\right]\right.\\ &=\tan ^{-1}\left[2 \sin \left(\cos ^{-1} \frac{1}{2}\right)\right] = \frac{\pi}{3}\end{aligned}

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