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Provide solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise  3.7 question 3 sub question (iv)

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Answer:   \frac{\pi }{4}

Hint:  The range of principal value of \tan ^{-1} is \left [-\frac{\pi }{2} ,\frac{\pi }{2}\right ]
Given: \tan ^{-1}\left(\tan \frac{9 \pi}{4}\right)


First we solve \tan \frac{9 \pi}{4}

            \tan \frac{9 \pi}{4}=\tan \left(2 \pi+\frac{\pi}{4}\right)

            \therefore \tan (2 \pi+\theta)=\tan \theta

            \begin{aligned} &\tan \left(2 \pi+\frac{\pi}{4}\right)=\tan \frac{\pi}{4} \\ &\therefore \tan \frac{\pi}{4}=1 \\ &\tan \frac{9 \pi}{4}=1 \end{aligned}

By substituting this value in \tan ^{-1}\left(\tan \frac{9 \pi}{4}\right), we get

            \tan ^{-1}(1)

let,        \tan ^{-1}(1)=y

            \begin{aligned} &\tan y=1 \\ &\tan \left(\frac{\pi}{4}\right)=1 \end{aligned}

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

            \tan ^{-1}\left(\tan \frac{9 \pi}{4}\right)=\frac{\pi}{4}\; \; \; \; \; \; \; \; \quad \tan ^{-1}(\tan x)=x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

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