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

$\begin{array}{c}\mathrm{f}\left(\mathrm{n\theta }\right)=\frac{2\sin 2\mathrm{\theta }}{\cos 2\mathrm{\theta }-\cos 4\mathrm{n\theta }}\\ =\frac{2\sin 2\mathrm{\theta }}{2\sin (2\mathrm{n}+1)\mathrm{\theta sin}(2\mathrm{n}-1)\mathrm{\theta }}\\ =\frac{\sin \left(\right(2\mathrm{n}+1)\mathrm{\theta }-(2\mathrm{n}-1\left)\mathrm{\theta }\right)}{\sin (2\mathrm{n}+1)\mathrm{\theta sin}(2\mathrm{n}-1)\mathrm{\theta }}\\ =\frac{\sin (2\mathrm{n}+1)\mathrm{\theta cos}(2\mathrm{n}-1)\mathrm{\theta }-\cos (2\mathrm{n}+1)\mathrm{\theta sin}(2\mathrm{n}-1)\mathrm{\theta }}{\sin (2\mathrm{n}+1)\mathrm{\theta sin}(2\mathrm{n}-1)\mathrm{\theta }}\\ \begin{array}{cc} & =\cot (2\mathrm{n}-1)\mathrm{\theta }-\cot (2\mathrm{n}+1)\mathrm{\theta }\\ \therefore & \mathrm{f}\left(\mathrm{\theta }\right)+\mathrm{f}\left(2\mathrm{\theta }\right)+\mathrm{f}\left(3\mathrm{\theta }\right)+\dots +\mathrm{f}\left(\mathrm{n\theta }\right)\\ & =\cot \mathrm{\theta }-\cot (2\mathrm{n}+1)\mathrm{\theta }\\ & =\frac{\sin (2\mathrm{n}+1)\mathrm{\theta cos}\mathrm{\theta }-\cos (2\mathrm{n}+1)\mathrm{\theta sin}\mathrm{\theta }}{\sin (2\mathrm{n}+1)\mathrm{\theta sin}\mathrm{\theta }}\\ & =\frac{\sin 2\mathrm{n\theta }}{\sin (2\mathrm{n}+1)\mathrm{\theta sin}\mathrm{\theta }}\end{array}\end{array}$