Question:
If $4 \tan \theta=3$, then $\left(\frac{4 \sin \theta-\cos \theta}{4 \sin \theta+\cos \theta}\right)$ is equal to
(a) $\frac{2}{3}$
(b) $\frac{1}{3}$
(c) $\frac{1}{2}$
(d) $\frac{3}{4}$
Solution:
(c) Given, $4 \tan \theta=3$
$\Rightarrow$ $\tan \theta=\frac{3}{4}$ ...(i)
$\therefore$ $\frac{4 \sin \theta-\cos \theta}{4 \sin \theta+\cos \theta}=\frac{4 \frac{\sin \theta}{\cos \theta}-1}{4 \frac{\sin \theta}{\cos \theta}+1}$
[divide by $\cos \theta$ in both numerator and denominator]
$=\frac{4 \tan \theta-1}{4 \tan \theta+1}$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
$=\frac{4\left(\frac{3}{4}\right)-1}{4\left(\frac{3}{4}\right)+1}=\frac{3-1}{3+1}=\frac{2}{4}=\frac{1}{2}$ [put the value from Eq. (i)]