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need solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise 3.7 question 5 sub question (v)

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Answer:   \frac{\pi }{6}

Hint: The range of principal value of    \operatorname{cosec}^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}
Given:  \operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{13 \pi}{6}\right)


First we solve 

            \operatorname{cosec}\left(\frac{13 \pi}{6}\right)=\operatorname{cosec}\left(2 \pi+\frac{\pi}{6}\right)

As we know   \operatorname{cosec}(2 \pi+\theta)=\operatorname{cosec} \theta

            \csc (2\pi+\frac{\pi}{6})=\csc (\frac{\pi}{6})=2

By substituting these value in \operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{13 \pi}{6}\right) we get,


Now,     \text { let } y=\operatorname{cosec}^{-1}(2)

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

The range of the principal value in \operatorname{cosec}^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\} \text { and } \operatorname{cosec}\left(\frac{\pi}{6}\right)=2

            \begin{aligned} &\therefore \operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{13 \pi}{6}\right) \text { is } \frac{\pi}{6} \\ &\therefore \operatorname{cosec}^{-1}(\operatorname{cosec} x)=x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\} \end{aligned}

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