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

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Answer: 2-\pi

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


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

But here x=2  which does not belong to above range

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

Now,         2-\pi in the given range \left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]

Hence,      \tan ^{-1}(\tan 2)=2-\pi

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