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explain solution RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 3 sub question (ii) maths

Answers (1)

Answer:   -\frac{\pi }{7}

Hint:  The range of principal value of \tan ^{-1} is \left [-\frac{\pi }{2} ,\frac{\pi }{2}\right ]
Given: \tan ^{-1}\left(\tan \frac{6 \pi}{7}\right)

Explanation:

First we solve \tan \frac{6\pi }{7}

            \begin{aligned} &\tan \frac{6 \pi}{7}=\tan \left(\pi-\frac{\pi}{7}\right) \\ &\tan \left(\pi-\frac{\pi}{7}\right)=-\tan \frac{\pi}{7}\; \; \; \; \; \; \; \; \; \quad(\tan (\pi-\theta)=-\tan \theta) \end{aligned}

We know 

            \tan ^{-1}(\tan x)=x \text { if } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

            \tan ^{-1}\left(\tan \frac{6 \pi}{7}\right)=-\frac{\pi}{7}

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