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provide solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise 3.7 question 1 sub question (x)

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Answer:   \pi -2

Hint: The principal value branch of function  \sin ^{-1}  is  \left [ -\frac{\pi }{2},\frac{\pi }{2} \right ]

Given:   \sin ^{-1}(\sin 2)


We know that \sin ^{-1}(\sin x)=x \text { with } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] which is approximately equal to[-1.57, 1.57]

But here x=2 , which do not lie on the above range,

we know that   \sin (\pi-x)=\sin (x)

                \begin{aligned} &\sin (\pi-2)=\sin (2) \text { also } \pi-2 \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\ &\sin ^{-1}(\sin 2)=\sin ^{-1}(\sin \pi-2) \\ &\therefore \sin ^{-1}(\sin \pi-2)=\pi-2 \end{aligned}

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