if

Let $\cot ^{-1} 3=\theta \Rightarrow \cot \theta=3$

$\therefore \cot \left(\frac{\pi}{4}-2 \cot ^{-1} 3\right)$

$=\cot \left(\frac{\pi}{4}-2 \theta\right)$

$=\frac{\cot \frac{\pi}{4} \cot 2 \theta+1}{\cot 2 \theta-\cot \frac{\pi}{4}}$                        $\left[\cot (A-B)=\frac{\cot A \cot B+1}{\cot B-\cot A}\right]$

$=\frac{\cot 2 \theta+1}{\cot 2 \theta-1}$

$=\frac{\frac{\cot ^{2} \theta-1}{2 \cot \theta}+1}{\frac{\cot ^{2} \theta-1}{2 \cot \theta}-1}$

$=\frac{\cot ^{2} \theta-1+2 \cot \theta}{\cot ^{2} \theta-1-2 \cot \theta}$

$=\frac{3^{2}-1+2 \times 3}{3^{2}-1-2 \times 3}$                      $(\cot \theta=3)$

$=\frac{9-1+6}{9-1-6}$

$=\frac{14}{2}$

$=7$

$\cot \left(\frac{\pi}{4}-2 \cot ^{-1} 3\right)$ is equal to $7 .$