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explain solution RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 6 sub question (ii) maths

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Answer:    \frac{\pi }{3}

Hint: The range of principal value of    \cot ^{-1}x is \left [ 0,\pi \right ]
Given:  \cot ^{-1}\left(\cot \frac{4\pi}{3}\right)


First we solve \cot \frac{4 \pi}{3}

            \cot \left(\frac{4 \pi}{3}\right)=\cot \left(\pi+\frac{\pi}{3}\right)

As we know   \cot (\pi+\theta)=\cot \theta

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

By substituting these value \cot ^{-1}\left(\cot \frac{4 \pi}{3}\right) we get,

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

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

            \begin{aligned} &\cot y=\frac{\sqrt{3}}{3} \\ &\cot \left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{3} \end{aligned}

The range of the principal value of   \cot ^{-1} \text { is }[0, \pi]

            \begin{aligned} &\cot ^{-1}\left(\cot \frac{4 \pi}{3}\right)=\frac{\pi}{3} \\ &\therefore \cot ^{-1}(\cot x)=x, x \in[0, \pi] \end{aligned}

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