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  • Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise  Very short answer question 11 math

Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise  Very short answer question 11 math

Answers (1)

Answer: \frac{2 \pi}{3}

Given:

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

Hint: The range of \sin is \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right), \ \ range \ \ of \ \ \cos is \ \ \((0, \pi) 

Solution:

            

\begin{aligned} \cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right) &=\cos ^{-1}\left(\cos \frac{\pi}{3}\right)+2 \sin ^{-1}\left(\sin \frac{\pi}{6}\right) \\ \\&=\frac{\pi}{3}+2\left(\frac{\pi}{6}\right) \\\\ &=\frac{\pi}{3}+\frac{\pi}{3} \\ \\&=\frac{2 \pi}{3} \end{aligned}

 

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