# 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) $\frac{3}{4}$

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Given:

$4 \tan \theta=3$

To do:

We have to find the value of $\left(\frac{4 \sin \theta-\cos \theta}{4 \sin \theta+\cos \theta}\right)$.

Solution:

$4 \tan \theta=3$

$\tan \theta=\frac{3}{4}$

Therefore,

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

Dividing both numerator and denominator by $\cos \theta$, we get,

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

$=\frac{4 \tan \theta-1}{4 \tan \theta+1}$          [Since $\tan \theta=\frac{\sin \theta}{\cos \theta}$]

$=\frac{4(\frac{3}{4})-1}{4(\frac{3}{4})+1}$           ($\tan \theta=\frac{3}{4}$)

$=\frac{3-1}{3+1}$

$=\frac{2}{4}$

$=\frac{1}{2}$

Updated on 10-Oct-2022 13:28:58