If $3\tan\theta = 4$, then evaluate $\frac{3sin\theta + 2cos\theta}{3sin\theta - 2cos\theta}$.

Explanation

$3\tan\theta = 4$ $\tan\theta = \frac{4}{3}$ From Pythagoras Theorem $AC^2 = AB^2 + BC^2$ $AC^2 = 4^2 + 3^2$ $AC^2 = 16 + 9$ $AC^2 = 25$ $AC = 5$ $\therefore sin\theta = \frac{4}{5}, cos\theta = \frac{3}{5}$ $\therefore \frac{3sin\theta + 2cos\theta}{3sin\theta - 2cos\theta} = \frac{3 \times \frac{4}{5} + 2 \times \frac{3}{5}}{3 \times \frac{4}{5} - 2 \times \frac{3}{5}}$ $= \frac{\frac{12}{5} + \frac{6}{5}}{\frac{12}{5} - \frac{6}{5}} = \frac{18}{5} \times \frac{5}{6} = 3$