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Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise  Very short answer question 51 math

Answers (1)

Answer:  \frac{\pi }{3}

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

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

Solution:

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

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