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Provide solution for RD Sharma math class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 2

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Answer: \pi

Given: \sin ^{-1} x, x \in(-1,1)

Hint: The maximum value of \sin ^{-1} x  in x \in (-1,1)  is at 1 .


               \sin ^{-1}(1)  maximum value is,

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


Again the minimum value is at -1


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

The difference between the minimum value, maximum value is


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