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  • explain solution RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 5 sub question (ii) maths

explain solution RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 5 sub question (ii) maths

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Answer: \frac{\pi }{4}

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{3 \pi}{4}\right)

Explanation:

First we solve \operatorname{cosec}\left(\frac{3 \pi}{4}\right)

            \operatorname{cosec}\left(\frac{3 \pi}{4}\right)=\operatorname{cosec}\left(\pi-\frac{\pi}{4}\right)

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

            \begin{aligned} &\operatorname{cosec}\left(\pi-\frac{\pi}{4}\right)=\operatorname{cosec}\left(\frac{\pi}{4}\right) \\ &\operatorname{cosec}\left(\frac{\pi}{4}\right)=\sqrt{2} \end{aligned}

By substituting these values in \operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{3 \pi}{4}\right) we get,

            \operatorname{cosec}^{-1}(\sqrt{2})

Let,        y=\operatorname{cosec}^{-1}(\sqrt{2})

            \begin{aligned} &\operatorname{cosec} y=\sqrt{2} \\ &\operatorname{cosec}\left(\frac{\pi}{4}\right)=\sqrt{2} \end{aligned}

The range of the principal value of   \operatorname{cosec}^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\} \text { and } \operatorname{cosec}\left(\frac{\pi}{4}\right)=\sqrt{2}

            \operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{\pi}{4}\right)=\frac{\pi}{4}

Hence,  \operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{3 \pi}{4}\right)=\frac{\pi}{4}

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