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Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 1 sub question (ix) maths textbook solution

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Answer:   12-4{\pi }

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

Given:   \sin ^{-1}(\sin 12)

Explanation:

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  x=12, which do not lie on the above range

We know  \sin (2 n \pi-x)=\sin (-x)

Hence   \sin (2 n \pi-12)=\sin (-12)

Here, 

        \begin{aligned} &n=2 \text { also } 12-4 \pi \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\ &\sin ^{-1}(\sin 12)=12-4 \pi \end{aligned}

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