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need solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise 3.7 question 6  sub question (i)

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Answer:     \frac{\pi }{3}

Hint: The range of principal value of    \cot ^{-1} is \left [ 0,\pi \right ]
Given:  \cot ^{-1}\left(\cot \frac{\pi}{3}\right)

Explanation:

As we know

                \cot ^{-1}(\cot x)=x \text { provided } \mathrm{} x \in[0, \pi]

By apply this situation we get,

                \cot ^{-1}\left(\cot \frac{\pi}{3}\right)=\frac{\pi}{3}

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