Given, $4 \tan \theta=3$

$\Rightarrow \quad \tan \theta=\frac{3}{4}$ ……….$(i)$

$\frac{4 \sin \theta-\cos \theta}{4 \sin \theta+\cos \theta}=\frac{4 \frac{\sin \theta}{\cos \theta}-1}{4 \frac{\sin \theta}{\cos \theta}+1}$

[divide by $\cos \theta$ in both numerator and denominator]

$\left.=\frac{4 \tan \theta-1}{4 \tan \theta+1} \quad [\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$

$=\frac{4\left(\frac{3}{4}\right)-1}{4\left(\frac{3}{4}\right)+1}=\frac{3-1}{3+1}=\frac{2}{4}=\frac{1}{2} .$ [put the value from Eq. $(i)$ ]