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Provide Solution For R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonometric Functions  Exercise MCQs Question 19  Maths Textbook Solution.

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Answer: \frac{-\pi}{10}

Hint: Just separate cos function into two parts.

Given: \sin ^{-1}\left(\cos \frac{33 \pi}{5}\right)

We have to compute x.


\sin ^{-1}\left(\cos \frac{33 \pi}{5}\right)_{=} \sin ^{-1}\left\{\cos \left(6 \pi+\frac{3 \pi}{5}\right)\right\}

                                =\sin ^{-1}\left\{\sin \left(\frac{\pi}{2}-\frac{3 \pi}{5}\right)\right\}

                                \begin{aligned} &=\frac{\pi}{2}-\frac{3 \pi}{5} \\ &=\frac{-\pi}{10} \end{aligned}

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