448. $\frac{\sin 7 \alpha-\sin 5 \alpha}{\sin 7 \alpha+\sin 5 \alpha}$.
Solution. Applying formulas (20) and (21), we get

$$
\frac{\sin 7 \alpha-\sin 5 \alpha}{\sin 7 \alpha+\sin 5 \alpha}=\frac{2 \sin \frac{7 \alpha-5 \alpha}{2} \cos \frac{7 \alpha+5 \alpha}{2}}{2 \sin \frac{7 \alpha+5 \alpha}{2} \cos \frac{7 \alpha-5 \alpha}{2}}=\frac{\sin \alpha \cos 6 \alpha}{\sin 6 \alpha \cos \alpha}=\operatorname{tg} \alpha \operatorname{ctg} 6 \alpha
$$