Solve this

When $x \neq \frac{\pi}{4}, \tan \left(\frac{\pi}{4}-x\right)$ and $\cot 2 x$ are continuous in $\left[0, \frac{\pi}{2}\right]$.

Thus, the quotient function $\frac{\tan \left(\frac{\pi}{4}-x\right)}{\cot 2 x}$ is continuous in $\left[0, \frac{\pi}{2}\right]$ for each $x \neq \frac{\pi}{4}$.

So, if $f(x)$ is continuous at $x=\frac{\pi}{4}$, then it will be everywhere continuous in $\left[0, \frac{\pi}{2}\right]$.

Now,

Let us consider the point $x=\frac{\pi}{4}$.

Given: $f(x)=\frac{\tan \left(\frac{\pi}{4}-x\right)}{\cot (2 x)}, x \neq \frac{\pi}{4}$

We have

$\left(\mathrm{LHL}\right.$ at $\left.x=\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{4}-h\right)=\lim _{h \rightarrow 0}\left(\frac{\tan \left(\frac{\pi}{4}-\frac{\pi}{4}+h\right)}{\cot \left(\frac{\pi}{2}-2 h\right)}\right)=\lim _{h \rightarrow 0}\left(\frac{\tan (h)}{\tan (2 h)}\right)=\lim _{h \rightarrow 0}\left(\frac{\frac{\tan (h)}{h}}{\frac{2 \tan (2 h)}{2 h}}\right)=\frac{1}{2}\left(\frac{\lim _{h \rightarrow 0} \frac{\tan (h)}{h}}{\lim _{h \rightarrow 0} \frac{\tan (2 h)}{2 h}}\right)=\frac{1}{2}$

$\left(\mathrm{RHL}\right.$ at $\left.x=\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}^{+}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{4}+h\right)=\lim _{h \rightarrow 0}\left(\frac{\tan \left(\frac{\pi}{4}-\frac{\pi}{4}-h\right)}{\cot \left(\frac{\pi}{2}+2 h\right)}\right)=\lim _{h \rightarrow 0}\left(\frac{\tan (-h)}{-\tan (2 h)}\right)=\lim _{h \rightarrow 0}\left(\frac{\tan (h)}{\tan (2 h)}\right)=\lim _{h \rightarrow 0}\left(\frac{\frac{\tan (h)}{h}}{\frac{2 \tan (2 h)}{2 h}}\right)=\frac{1}{2}\left(\frac{\lim _{h \rightarrow 0} \frac{\tan (h)}{h}}{\lim _{h \rightarrow 0} \frac{\tan (2 h)}{2 h}}\right)=\frac{1}{2}$

If $f(x)$ is continuous at $x=\frac{\pi}{4}$, then

$\lim _{x \rightarrow \frac{\pi}{4}^{-}} f(x)=\lim _{x \rightarrow \frac{\pi}{4}^{+}} f(x)=f\left(\frac{\pi}{4}\right)$

$\therefore f\left(\frac{\pi}{4}\right)=\frac{1}{2}$

Hence, for $f\left(\frac{\pi}{4}\right)=\frac{1}{2}$, the function $f(x)$ will be everywhere continuous in $\left[0, \frac{\pi}{2}\right]$.