15. Let $0<\theta<\frac{\pi}{2}$, find the minimum value of $\frac{1}{\sin \theta}+\frac{3 \sqrt{3}}{\cos \theta}$, and the value of $\theta$ when the minimum value is obtained.
15. $4+\frac{1}{\sin \theta}+\frac{3 \sqrt{3}}{\cos \theta}=\frac{1}{\sin \theta}+4 \sin ^{2} \theta+\frac{3 \sqrt{3}}{\cos \theta}+4 \cos ^{2} \theta=\left(\frac{1}{2 \sin \theta}+\frac{1}{2 \sin \theta}+4 \sin ^{2} \theta\right)+$ $\left(\frac{3 \sqrt{3}}{2 \cos \theta}+\frac{3 \sqrt{3}}{2 \cos \theta}+4 \cos ^{2} \theta\right) \geqslant 3+3 \sqrt[3]{27}=12$, so $\frac{1}{\sin \theta}+\frac{3 \sqrt{3}}{\cos \theta} \geqslant 8$, when and only when $\frac{1}{2 \sin \theta}=$ $4 \sin ^{2} \theta \frac{3 \sqrt{3}}{2 \cos \theta}=4 \cos ^{2} \theta$, and $\sin \theta=\frac{1}{2}, \cos \theta=\frac{\sqrt{3}}{2}$, i.e., $\theta=\frac{\pi}{6}$, the equality holds.