Given that $\mathrm{f}\left(\mathrm{n\theta }\right)=\frac{2\sin 2\mathrm{\theta }}{\cos 2\mathrm{\theta }-\cos 4\mathrm{n\theta }}$ and $\mathrm{f}\left(\mathrm{\theta }\right)+\mathrm{f}\left(2\mathrm{\theta }\right)+\mathrm{f}\left(3\mathrm{\theta }\right)+\dots +\mathrm{f}\left(\mathrm{n\theta }\right)=\frac{\sin \mathrm{\lambda \theta }}{\sin \mathrm{\theta sin}\mathrm{\mu \theta }}$ then the value of $\mathrm{\mu }-\mathrm{\lambda }$, is ____

$\begin{array}{l}\mathrm{f}\left(\mathrm{n\theta }\right)=\frac{2\sin 2\mathrm{\theta }}{\cos 2\mathrm{\theta }-\cos 4\mathrm{n\theta }}=\frac{2\sin 2\mathrm{\theta }}{2\sin (2\mathrm{n}+1)\mathrm{\theta sin}(2\mathrm{n}-1)\mathrm{\theta }}\\ =\cot (2\mathrm{n}-1)\mathrm{\theta }-\cot (2\mathrm{n}+1)\mathrm{\theta }\end{array}$

$$\begin{gathered} \sum _{r=1}^{n}f\left(r\theta \right)=cot\theta -cot\left(2n+1\right)\theta =\frac{\sin \lambda \theta }{\sin \theta \sin \mu \theta } \\ \frac{\cos \theta }{\sin \theta }-\frac{\cos \left(2n+1\right)\theta }{\sin \left(2n+1\right)\theta }=\frac{\sin \lambda \theta }{\sin \theta \sin \mu \theta } \\ \frac{\sin 2n\theta }{\sin \theta \sin \left(2n+1\right)\theta }=\frac{\sin \lambda \theta }{\sin \theta \sin \mu \theta }\Rightarrow \lambda =2n,\mu =2n+1\Rightarrow \mu -\lambda =1 \end{gathered}$$