If $tan\theta =\frac{4}{5}$ then find the value of $\frac{5sin\theta -3cos\theta }{5sin\theta +3cos\theta }$.

Given $\tan \theta =\frac{4}{5}$

$\frac{5\sin \theta -3\cos \theta }{5\sin \theta +3\cos \theta }=\frac{5\left(\frac{\sin \theta }{\cos \theta }\right)-3}{5\left(\frac{\sin \theta }{\cos \theta }\right)+3}=\frac{5\times \frac{4}{5}-3}{5\times \frac{4}{5}+3}=\frac{4-3}{4+3}=\frac{1}{7}$