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Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 3 sub question (iii) maths textbook solution

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Answer:   \frac{\pi }{6}

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


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

            \tan \frac{7 \pi}{6}=\tan \left(\pi+\frac{\pi}{6}\right)

As we know, \tan (\pi+\theta)=\tan \theta

            \begin{aligned} &\tan \left(\pi+\frac{\pi}{6}\right)=\tan \frac{\pi}{6} \\ &\therefore \tan \frac{\pi}{6}=\frac{1}{\sqrt{3}} \end{aligned}

By substituting this value in \tan ^{-1}\left(\tan \frac{7 \pi}{6}\right) we get

            \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)

Now,     \text { let } \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=y

            \begin{aligned} &\tan y=\frac{1}{\sqrt{3}} \\ &\tan \left(\frac{\pi}{6}\right)=\frac{1}{\sqrt{3}} \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}{6}\right)=\frac{1}{\sqrt{3}}

            \tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)=\frac{\pi}{6} \quad \text { As } \quad \tan ^{-1}(\tan x)=x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

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