Question : If $3 \tan \theta=2$, what is the value of $\frac{(3 \sin \theta-\cos \theta)}{(3 \sin \theta+\cos \theta)}?$
Option 1: $1$
Option 2: $3$
Option 3: $\sqrt3$
Option 4: $\frac{1}{3}$
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Correct Answer: $\frac{1}{3}$
Solution :
$3 \tan \theta = 2$
$⇒\tan \theta = \frac{2}{3}$
Now, $\frac{(3 \sin \theta-\cos \theta)}{(3 \sin \theta+\cos \theta)}$
Dividing numerator and denominator by $\cos \theta$, we get,
$=\frac{(3\tan \theta - 1)}{(3\tan \theta+1)}$
$=\frac{3 \times \frac{2}{3} - 1}{3\times \frac{2}{3}+1}$
$=\frac{1}{3}$
Hence, the correct answer is $\frac{1}{3}$.
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