Using the small angle approximations:

$\frac{1-\cos(3\theta)}{\theta\tan(2\theta)}\approx \frac{1-\left(1-\frac{(3\theta)^2}{2}\right)}{\theta\times 2\theta}=\frac{\frac{9\theta^2}{2}}{2\theta^2}=\frac{9}{4}$

Hence, for $\frac{1-\cos(3\theta)}{\theta\tan(2\theta)}$ is approximately $\frac{9}{4}$ for $\theta$ close to 0.