If $f(x)=\left\{\begin{array}{cc}\frac{\tan \left(\frac{\pi}{4}-x\right)}{\cot 2 x}, & x \neq \frac{\pi}{4} \\ k, & x=\frac{\pi}{4}\end{array}\right.$ is continuous at $x=\frac{\pi}{4}$, then the value of 'k' is:

$\frac{1}{2}$

The correct answer is Option (3) - $\frac{1}{2}$

$f(\frac{\pi}{4})=k$

$\lim\limits_{x→\frac{\pi}{4}}\left(\frac{(\frac{\pi}{4}-x)}{\cot 2x}=\frac{1-\tan x}{1+\tan x}×\tan 2x\right)$

$\lim\limits_{x→\frac{\pi}{4}}\left(\frac{1-\tan x}{1+\tan x}×\frac{2\tan x}{1+\tan x}=\frac{2\tan x}{(1+\tan x)^2}\right)$

$=\frac{2×1}{2^2}=\frac{1}{2}=k$