Question

If $4\tan$$\theta$$=3$, then evaluate $\frac{(4\sin \theta -\cos \theta )}{(4\sin \theta +\cos \theta )}$

Answer 1

Given $4Tan\theta =3\Rightarrow Tan\theta =\frac{3}{4}$

$sin\theta =\frac{3}{5}cos\theta =\frac{4}{5}$

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

$=\frac{\frac{8}{5}}{\frac{16}{5}}=\frac{1}{2}$