If $f(x)=\left\{\begin{matrix}\frac{\tan\left(\frac{π}{4}-x\right)}{\cot 2x},&x≠\frac{π}{4}\\k&,x=\frac{π}{4}\end{matrix}\right.$ is continuous at $x=\frac{π}{4}$, then the value of $k$ will be equal to

$\frac{1}{2}$

$f(x)=\left\{\begin{matrix}\frac{\tan\left(\frac{π}{4}-x\right)}{\cot 2x},&x≠\frac{π}{4}\\k&,x=\frac{π}{4}\end{matrix}\right.$

$f(\frac{π}{4})=k\underset{x→\frac{π}{4}}{\lim}\frac{\tan\left(\frac{π}{4}-x\right)}{\cot 2x}$ → $\frac{0}{0}$ form

Using L'Hopital rule

$\underset{x→\frac{π}{4}}{\lim}\frac{-\sec^2\left(\frac{π}{4}-x\right)}{-2\,cosec^2\,2x}$

$=\frac{\sec^2(0)}{2\,cosec^2\,\frac{π}{2}}$

$=\frac{1}{2}=k$