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  • need solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise 3.7 question 1 sub question (iii)

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

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Answer:    \frac{\pi }{6}

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

Given:   \sin ^{-1}\left(\sin \frac{5 \pi}{6}\right)

Explanation:

First, we solve  \sin \frac{5\pi }{6}

        \frac{5 \pi}{6}=\pi-\frac{\pi}{6}

Then,

         \begin{aligned} &\sin \left(\pi-\frac{\pi}{6}\right)=\sin \left(\frac{\pi}{6}\right) \\ &\sin (\pi-\theta)=\sin (\theta) \end{aligned}

We know that the value of  \sin \frac{\pi }{6}  is \frac{1}{2}

            \sin \frac{5 \pi}{6}=\frac{1}{2}

By substituting this value in  \sin ^{-1}\left(\sin \frac{5 \pi}{6}\right)

we get     \sin ^{-1}\left(\frac{1}{2}\right)

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

                \begin{aligned} &\sin y=\frac{1}{2} \\ &\sin \left(\frac{\pi}{6}\right)=\frac{1}{2} \end{aligned}

The range of principal value of  \sin ^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \text { and } \sin \left(\frac{\pi}{6}\right)=\frac{1}{2}

                \begin{aligned} &\sin ^{-1}\left(\sin \frac{\pi}{6}\right)=\frac{\pi}{6} \\ &\sin ^{-1}\left(\sin \frac{5 \pi}{6}\right)=\frac{\pi}{6} \quad \because\left[\sin ^{-1}(\sin \theta)=\theta\right] \end{aligned}

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