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  • Explain solution for RD Sharma class class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 8 math

Explain solution for RD Sharma class class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 8 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:

Range of  \sin \left(\frac{-\pi}{2}, \frac{\pi}{2}\right) ; \frac{\pi}{3} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)

Range of    \cos [0, \pi], \frac{2 \pi}{3} \in(0, \pi)

Solution:

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

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