If \(\tan \theta = \frac{4}{3}\), then the value of \(\frac{3\sin \theta + 2\cos \theta}{3\sin \theta - 2\cos \theta}\) is:
Answer: B
We can divide both the numerator and the denominator by \(\cos \theta\).
Expression = \(\frac{3(\frac{\sin \theta}{\cos \theta}) + 2(\frac{\cos \theta}{\cos \theta})}{3(\frac{\sin \theta}{\cos \theta}) - 2(\frac{\cos \theta}{\cos \theta})} = \frac{3\tan \theta + 2}{3\tan \theta - 2}\).
Substitute \(\tan \theta = 4/3\):
\(\frac{3(4/3) + 2}{3(4/3) - 2} = \frac{4+2}{4-2} = \frac{6}{2} = 3\).