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

explain solution RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 5 sub question (vi) 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}\left(-\frac{9 \pi}{4}\right)\right\}

Explanation:

First we solve  \operatorname{cosec}\left(-\frac{9 \pi}{4}\right)

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

                     \begin{aligned} &\operatorname{cosec}\left(-\frac{9 \pi}{4}\right)=-\operatorname{cosec}\left(\frac{9 \pi}{4}\right) \\ &-\operatorname{cosec}\left(\frac{9 \pi}{4}\right)=-\operatorname{cosec}\left(2 \pi+\frac{\pi}{4}\right) \\ &-\operatorname{cosec}\left(2 \pi+\frac{\pi}{4}\right)=-\operatorname{cosec}\left(\frac{\pi}{4}\right) \end{aligned}

As we know

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

Now it becomes   \operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{-\pi}{4}\right)

                    \begin{aligned} &\therefore \operatorname{cosec}^{-1}(\operatorname{cosec} x)=x \text { provide } \mathrm{d} x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\} \\ &\operatorname{cosec}^{-1}\left(\operatorname{cosec}\left(\frac{\pi}{4}\right)\right)=-\frac{\pi}{4} \end{aligned}

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