It is given that $4 \tan \mathrm{\theta }=3$ and we need to find the value of $\left(\frac{4\mathrm{sin\theta }-\mathrm{cos\theta }}{4\mathrm{sin\theta }+\mathrm{cos\theta }}\right)$.

Let $\left(\frac{4\mathrm{sin\theta }-\mathrm{cos\theta }}{4\mathrm{sin\theta }+\mathrm{cos\theta }}\right)$.........(i)

And $4\tan \mathrm{\theta }=3 ..............\left(\mathrm{ii}\right)$

Equation (i) can be also be written as $\tan \mathrm{\theta }=\frac{3}{4}$.

On dividing both sides of (i) with $\cos \mathrm{\theta }$, we get $\left(\frac{4\mathrm{tan\theta }-1}{4\mathrm{tan\theta }+1}\right) .............\left(\mathrm{iii}\right)$                                                                                                                  

On substituting the value of $\mathrm{\theta }$ in (iii), we get

$\begin{array}{l}\Rightarrow \left(\frac{4\mathrm{tan\theta }-1}{4\mathrm{tan\theta }+1}\right)=\frac{4\left(\frac{3}{4}\right)-1}{4\left(\frac{3}{4}\right)+1}\\ \Rightarrow \left(\frac{4\mathrm{tan\theta }-1}{4\mathrm{tan\theta }+1}\right)=\frac{3-1}{3+1}\\ \Rightarrow \left(\frac{4\mathrm{tan\theta }-1}{4\mathrm{tan\theta }+1}\right)=\frac{2}{4}\\ \Rightarrow \left(\frac{4\mathrm{tan\theta }-1}{4\mathrm{tan\theta }+1}\right)=\frac{1}{2}\end{array}$

Hence, the value of is  $\left(\frac{4\sin \theta -\cos \theta }{4\sin \theta +\cos \theta }\right)\text{ is }\frac{1}{2}$