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

Answer: $\frac{\pi }{3}$

Hint: The range of principal value of    $\cot ^{-1}$ is $\left [ 0,\pi \right ]$
Given:  $\cot ^{-1}\left\{\cot \left(-\frac{8 \pi}{3}\right)\right\}$

Explanation:

First we solve  $\cot \left(-\frac{8 \pi}{3}\right)$

As we know  $\cot (-\theta)=-\cot \theta$

\begin{aligned} &\cot \left(-\frac{8 \pi}{3}\right)=-\cot \left(\frac{8 \pi}{3}\right) \\ &-\cot \left(\frac{8 \pi}{3}\right)=-\cot \left(3 \pi-\frac{\pi}{3}\right) \end{aligned}

As we know, $\cot (2 \pi-\theta)=-\cot \theta$

\begin{aligned} -\cot \left(3 \pi-\frac{\pi}{3}\right) &=-\cot \left(-\frac{\pi}{3}\right) \\ &=\cot \left(\frac{\pi}{3}\right) \\ \cot \left(\frac{\pi}{3}\right) &=\frac{1}{\sqrt{3}} \end{aligned}

By substituting these values in $\cot ^{-1}\left(\cot \left(-\frac{8 \pi}{3}\right)\right)$ we get,

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

Now,

Let       $y=\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)$

\begin{aligned} &\cot y=\frac{1}{\sqrt{3}} \\ &\cot \left(\frac{\pi}{3}\right)=\frac{1}{\sqrt{3}} \end{aligned}

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

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