Question : If $\tan \theta=\frac{4}{3}$, then the value of $\frac{3\sin \theta+ 2\cos \theta}{3\sin \theta – 2 \cos \theta}$ is:
Option 1: $\frac{1}{2}$
Option 2: $1\frac{1}{2}$
Option 3: $3$
Option 4: $–3$
Correct Answer: $3$
Solution :
$\frac{3\sin\theta+2\cos\theta}{3\sin\theta–2\cos\theta}$
Divide both sides by $\cos\theta$,
= $\frac{\frac{3\sin\theta}{\cos\theta}+2}{\frac{3\sin\theta}{\cos\theta}–2}$
= $\frac{3\tan\theta+2}{3\tan\theta–2}$
= $\frac{3×\frac{4}{3}+2}{3×\frac{4}{3}-2}$
= $\frac{6}{2}$
= 3
Hence, the correct answer is $3$.
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