Given $\sin \left(\frac{\pi }{4}\mathrm{cot\theta }\right)=\cos \left(\frac{\pi }{4}\mathrm{tan\theta }\right)$

Using $\sin \left(\frac{\mathrm{\pi }}{2}-\theta \right)=\cos \theta ,$

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

$\Rightarrow \frac{\pi }{4}\mathrm{cot\theta }=\frac{\pi }{2}-\frac{\pi }{4}\mathrm{tan\theta }$

$\Rightarrow \frac{\pi }{4}\left(\mathrm{tan\theta }+\mathrm{cot\theta }\right)=\frac{\pi }{2}$

$\Rightarrow \mathrm{tan\theta }+\mathrm{cot\theta }=2$

$\Rightarrow \frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }=2$

$\Rightarrow \frac{{\sin }^2\theta +{\cos }^2\theta }{\cos \theta \sin \theta }=2$

$\Rightarrow \frac{1}{\mathrm{sin\theta cos\theta }}=2$

$\Rightarrow 2\mathrm{sin\theta cos\theta }=1$

Using $2\mathrm{sin\theta cos\theta }=\sin 2\theta ,$

$\Rightarrow \sin 2\mathrm{\theta }=1$

The solution of $\sin x=1$ is $x=2n\mathrm{\pi }+\frac{\mathrm{\pi }}{2}, n\in Z.$

$\Rightarrow 2\mathrm{\theta }=2n\pi +\frac{\pi }{2}, n\in Z$

$\Rightarrow \theta =n\pi +\frac{\pi }{4}.$

Hence, the solution of the given equation is $n\mathrm{\pi }+\frac{\mathrm{\pi }}{4}.$