449. $\frac{\cos \alpha - \cos 3 \alpha + \cos 5 \alpha - \cos 7 \alpha}{\sin \alpha + \sin 3 \alpha + \sin 5 \alpha + \sin 7 \alpha}$.
Solution. Grouping the terms and applying formulas (23) and (20), we factor out the common factors and simplify the fraction:

$$
\begin{gathered}
\frac{(\cos \alpha-\cos 3 \alpha)+(\cos 5 \alpha-\cos 7 \alpha)}{(\sin \alpha+\sin 3 \alpha)+(\sin 5 \alpha+\sin 7 \alpha)}= \\
=\frac{\left(-2 \sin \frac{\alpha-3 \alpha}{2} \sin \frac{\alpha+3 \alpha}{2}\right)+\left(-2 \sin \frac{5 \alpha-7 \alpha}{2} \sin \frac{5 \alpha+7 \alpha}{2}\right)}{2 \sin \frac{\alpha+3 \alpha}{2} \cos \frac{\alpha-3 \alpha}{2}+2 \sin \frac{5 \alpha+7 \alpha}{2} \cos \frac{5 \alpha-7 \alpha}{2}}= \\
=\frac{2 \sin \alpha \sin 2 \alpha+2 \sin \alpha \sin 6 \alpha}{2 \sin 2 \alpha \cos \alpha+2 \sin 6 \alpha \cos \alpha}=\frac{2 \sin \alpha(\sin 2 \alpha+\sin 6 \alpha)}{2 \cos \alpha(\sin 2 \alpha+\sin 6 \alpha)}=\frac{\sin \alpha}{\cos \alpha}=\operatorname{tg} \alpha .
\end{gathered}
$$