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  • Explain solution for RD Sharma class class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 40 math

Explain solution for RD Sharma class class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 40 math

Answers (1)

Answer: \pi

Given:

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

Hint:

\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)  is the range of the principal value branch of sine range of \cos (0, \pi)

Solution:

We have,

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

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