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

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Answer: 12-4\pi

Hint: The range of principal value of  \tan ^{-1} is \left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]
Given:  \tan ^{-1}(\tan 12)


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

But here x=12  which does not belongs to above range

            \begin{aligned} &\tan (2 n \pi-\theta)=-\tan (\theta) \\ &\tan (\theta-2 n \pi)=\tan \theta \end{aligned}

Here,     n=2

            \tan (12-4 \pi)=\tan (12)

Now,     12-4 \pi \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

Hence,  \tan ^{-1}(\tan 12)=12-4 \pi

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