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

Answer: $\frac{\pi }{4}$

Hint: The range of principal value of    $\cot ^{-1}$ is $\left [ 0,\pi \right ]$
Given:  $\cot ^{-1}\left\{\cot \left(\frac{21 \pi}{4}\right)\right\}$

Explanation:

First we solve $\cot \left(\frac{21 \pi}{4}\right)$

$\cot \left(\frac{21 \pi}{4}\right)=\cot \left(5 \pi+\frac{\pi}{4}\right)$

As we know  $\cot (n \pi+\theta)=\cot \theta \text { where } n \text { is }(1,2,3, \ldots)$

$\cot \left(5 \pi+\frac{\pi}{4}\right)=\cot \left(\frac{\pi}{4}\right) \quad\left[\because \cot \frac{\pi}{4}=1\right]$

$\cot \left(\frac{\pi}{4}\right)=1$

By substituting these values in  $\cot ^{-1}\left\{\cot \left(\frac{21 \pi}{4}\right)\right\}$  we get,

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

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

\begin{aligned} &\cot y=1 \\ &\cot \left(\frac{\pi}{4}\right)=1 \end{aligned}

The range of principal value of    $\cot ^{-1} \text { is }[0, \pi] \text { and } \cot \left(\frac{\pi}{4}\right)=1$

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