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  • Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 5 sub question (iii) maths textbook solution

Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 5 sub question (iii) maths textbook solution

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Answer:    -\frac{\pi }{5}

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{6 \pi}{5}\right)

Explanation:

First we solve \operatorname{cosec}\left(\frac{6 \pi}{5}\right)

            \operatorname{cosec}\left(\frac{6 \pi}{5}\right)=\operatorname{cosec}\left(\pi+\frac{\pi}{5}\right)

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

Then,     \operatorname{cosec}\left(\pi+\frac{\pi}{5}\right)=-\operatorname{cosec}\left(\frac{\pi}{5}\right)

            -\operatorname{cosec}\left(\frac{\pi}{5}\right)=\operatorname{cosec}\left(-\frac{\pi}{5}\right)

As         \operatorname{cosec}(-x)=-\operatorname{cosec} x

By substituting this value in \operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{6 \pi}{5}\right) we get,

             \operatorname{cosec}^{-1}\left(\operatorname{cosec}\left(-\frac{\pi}{5}\right)\right)

As we know,   \operatorname{cosec}^{-1}(\operatorname{cosec} x)=x \text { when } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}

            \operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{6 \pi}{5}\right)=-\frac{\pi}{5}

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