14. 设 $\alpha$ 为第四象限角, 若 $\frac{\sin 3 \alpha}{\sin \alpha}=\frac{13}{5}$, 则 $\tan 2 \alpha=$
解析 : $\frac{\sin 3 \alpha}{\sin \alpha}=\frac{\sin (2 \alpha+\alpha)}{\sin \alpha}=\frac{\sin 2 \alpha \cos \alpha+\cos 2 \alpha \sin \alpha}{\sin \alpha}=2 \cos ^{2} \alpha+\cos 2 \alpha=\frac{13}{5}$.

$\therefore 2 \cos ^{2} \alpha+2 \cos ^{2} \alpha-1=\frac{13}{5}$ ，

$\therefore \cos ^{2} \alpha=\frac{9}{10}, \sin ^{2} \alpha=\frac{1}{10}$ ，

$\therefore \tan ^{2} \alpha=\frac{\sin ^{2} \alpha}{\cos ^{2} \alpha}=\frac{1}{9}$,

$\because \alpha$ 为第四象限角， $\tan \alpha<0$.

$\therefore \tan \alpha=-\frac{1}{3}$.

$\therefore \tan 2 \alpha=\frac{2 \tan \alpha}{1-\tan ^{2} \alpha}=\frac{-\frac{2}{3}}{1-\frac{1}{9}}=-\frac{3}{4}$.