If $5\tan \mathrm{\theta }=4,\text{ then }\frac{5\sin \mathrm{\theta }-3\cos \mathrm{\theta }}{5\sin \mathrm{\theta }+2\cos \mathrm{\theta }}$ is equal to

$5\tan \mathrm{\theta }=4\text{ or }\tan \mathrm{\theta }=\frac{4}{5}$

,$\begin{array}{r}\mathrm{now}\frac{5\sin \mathrm{\theta }-3\cos \mathrm{\theta }}{5\sin \mathrm{\theta }+2\cos \mathrm{\theta }}=\frac{5\frac{\sin \mathrm{\theta }}{\cos \mathrm{\theta }}-3}{5\frac{\sin \mathrm{\theta }}{\cos \mathrm{\theta }}+2}\\ =\frac{5\tan \mathrm{\theta }-3}{5\tan \mathrm{\theta }+2}\\ =\frac{5\times \frac{4}{5}-3}{5\times \frac{4}{5}+2}=\frac{1}{6}\end{array}$