Given that,$\mathrm{f}\left(\mathrm{n\theta }\right)=\frac{2\sin 2\mathrm{\theta }}{\cos 2\mathrm{\theta }-\cos 3\mathrm{n\theta }}$,and  $\mathrm{f}\left(\mathrm{\theta }\right)+\mathrm{f}\left(2\mathrm{\theta }\right)+\mathrm{f}\left(3\mathrm{\theta }\right)+\dots +\mathrm{f}\left(\mathrm{n\theta }\right)=\frac{\sin \mathrm{\lambda \theta }}{\sin \mathrm{\theta sin}\mathrm{\mu \theta }}\text{, then }$the value of $\mathrm{\mu }-\mathrm{\lambda },\text{ is }$____.

Given that,

$\begin{array}{l}\mathrm{f}\left(\mathrm{n\theta }\right)=\frac{2\sin 2\mathrm{\theta }}{\cos 2\mathrm{\theta }-\cos 4\mathrm{n\theta }}\\ =\frac{2\sin 2\mathrm{\theta }}{2\sin (2\mathrm{n}+1)\mathrm{\theta sin}(2\mathrm{n}-1)\mathrm{\theta }}\\ =\frac{\sin (2\mathrm{n}+1)\mathrm{\theta }-(2\mathrm{n}-1)\mathrm{\theta }}{\sin (2\mathrm{n}+1)\mathrm{\theta sin}(2\mathrm{n}-1)\mathrm{\theta }}\\ =\frac{\sin (2\mathrm{n}+1)\mathrm{\theta cos}(2\mathrm{n}-1)\mathrm{\theta }-\cos (2\mathrm{n}+1)\mathrm{\theta sin}(2\mathrm{n}-1)\mathrm{\theta }}{\sin (2\mathrm{n}+1)\mathrm{\theta sin}(2\mathrm{n}-1)\mathrm{\theta }}\\ =\cot (2\mathrm{n}-1)\mathrm{\theta }-\cot (2\mathrm{n}+1)\mathrm{\theta }\\ \Rightarrow \mathrm{f}\left(\mathrm{\theta }\right)+\mathrm{f}\left(2\mathrm{\theta }\right)+\mathrm{f}\left(3\mathrm{\theta }\right)+\dots +\mathrm{f}\left(\mathrm{n\theta }\right)\\ =(\cot \mathrm{\theta }-\cot 3\mathrm{\theta })+(\cot 3\mathrm{\theta }-\cot 5\mathrm{\theta })+(\cot 5\mathrm{\theta }-\cot 7\mathrm{\theta })\\ =\cot \mathrm{\theta }-\cot (2\mathrm{n}+1)\mathrm{\theta }=\frac{\cos \mathrm{\theta }}{\sin \mathrm{\theta }}-\frac{\cos (2\mathrm{n}+1)\mathrm{\theta }}{\sin (2\mathrm{n}+1)\mathrm{\theta }}\\ =\frac{\sin 2\mathrm{n\theta }}{\sin \mathrm{\theta sin}(2\mathrm{n}+1)\mathrm{\theta }}\\ \Rightarrow \mathrm{\mu }=2\mathrm{n}+1\Rightarrow \mathrm{\lambda }=2\mathrm{n}\\ \Rightarrow \mathrm{\mu }-\mathrm{\lambda }=1\end{array}$