#### Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 6 sub question (iii) maths textbook solution

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

Hint: The range of principal value of    $\cot ^{-1}$ is $\left [ 0,\pi \right ]$
Given:  $\cot ^{-1}\left(\cot \frac{9\pi}{4}\right)$

Explanation:

First we solve  $\cot \frac{9 \pi}{4}$

\begin{aligned} &\cot \frac{9 \pi}{4}=\cot \left(2 \pi+\frac{\pi}{4}\right) \\ &\therefore \cot (2 \pi+\theta)=\cot \theta \\ &\cot \left(2 \pi+\frac{\pi}{4}\right)=\cot \frac{\pi}{4} \\ &\therefore \cot \frac{\pi}{4}=1 \end{aligned}

By substituting these value in $\cot ^{-1}\left(\cot \frac{9 \pi}{4}\right)$ 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}