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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: $\cot ^{-1}\left(\cot \frac{4\pi}{3}\right)$

Explanation:

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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