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

Answer: $\frac{\pi }{6}$

Hint: The range of principal value of    $\cot ^{-1}$ is $\left [ 0,\pi \right ]$
Given:  $\cot ^{-1}\left(\cot \frac{19\pi}{6}\right)$

Explanation:

First we solve  $\cot \left(\frac{19 \pi}{6}\right)$

$\cot \left(\frac{19 \pi}{6}\right)=\cot \left(3 \pi+\frac{\pi}{6}\right)$

As we know,  $\cot (n \pi+\theta)=\cot \theta$

\begin{aligned} &\cot \left(3 \pi+\frac{\pi}{6}\right)=\cot \frac{\pi}{6} \\ &\cot \frac{\pi}{6}=\sqrt{3} \end{aligned}                                $\left[\because \cot 30^{\circ}=\sqrt{3}\right]$

By substituting these value in $\cot ^{-1}\left(\cot \frac{19 \pi}{6}\right)$    we get,

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

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

\begin{aligned} &\cot y=\sqrt{3} \\ &\cot \left(\frac{\pi}{6}\right)=\sqrt{3} \end{aligned}

The range of the principal value of   $\cot ^{-1} \text { is }[0, \pi] \text { and } \cot \left(\frac{\pi}{6}\right)=\sqrt{3}$

$\therefore \cot ^{-1}\left(\cot \frac{\pi}{6}\right)=\frac{\pi}{6}$

As we know, $\cot ^{-1}(\cot x)=x, x \in[0, \pi]$