If 4 tan thita=3,evaluate 4 sin …

We have, 4 tan {tex}\theta{/tex} = 3
{tex}\Rightarrow{/tex} tan {tex}\theta{/tex} = {tex}\frac{3}{4}{/tex}

{tex}\because sec\theta=\sqrt{1+tan^2\theta}{/tex} = {tex}\sqrt{1+\frac{9}{16}}{/tex} ={tex}\sqrt{\frac{16+9}{16}} {/tex} {tex} = \sqrt{\frac{25}{16}}{/tex} {tex} = \frac{5}{4}{/tex}
Now,  {tex}\frac{4 \sin \theta-\cos \theta+1}{4 \sin \theta+\cos \theta-1}{/tex} 
= {tex}\frac{cos\theta(4 \frac{\sin \theta}{cos\theta}-\frac{\cos \theta}{cos\theta}+\frac{1}{cos\theta})}{cos\theta(4 \frac{\sin \theta}{{cos\theta}{}}+\frac{\cos \theta}{cos\theta}-\frac{1}{cos\theta})}{/tex}

= {tex}\frac{4tan\theta-1+sec\theta}{4tan\theta+1-sec\theta}{/tex}
Substituting the values, we get,
={tex}\frac{4(\frac{3}{4})-1+(\frac{5}{4})}{4(\frac{3}{4})+1-(\frac{5}{4})}{/tex} =  {tex}\frac{3-1+(\frac{5}{4})}{3+1-(\frac{5}{4})}{/tex} = {tex}\frac{2+(\frac{5}{4})}{4-(\frac{5}{4})}{/tex} = {tex}\frac{\frac{8+5}{4}}{\frac{16-5}{4}}{/tex} = {tex}\frac{\frac{13}{4}}{\frac{11}{4}}{/tex} = {tex}{\frac{13}{4}}\times{\frac{4}{11}}{/tex} = {tex}\frac{13}{11}{/tex}