$\cot(\frac{π}{4}-2\cot^{-1}3)$ is:

7

$\cot(\frac{π}{4}-2\cot^{-1}3)=\frac{1}{\tan(\frac{π}{4}-2\tan^{-1}\frac{1}{3})}$

$=\frac{1}{\tan(\frac{π}{4}-2\tan^{-1}\frac{3}{4})}$  [∵ $2\tan^{-1}\frac{1}{3}=\tan^{-1}(\frac{2×\frac{1}{3}}{1-\frac{1}{9}})=\tan^{-1}\frac{3}{4}$]

$=\frac{\tan\frac{π}{4}+\tan\tan^{-1}\frac{3}{4}}{1-\tan\frac{π}{4}\tan^{-1}\frac{3}{4}}$

$=\frac{1+\frac{3}{4}}{1-\frac{3}{4}}=7$