$7 \cdot 8 \frac{\sin 10^{\circ}+\sin 20^{\circ}}{\cos 10^{\circ}+\cos 20^{\circ}}$ can be simplified to
(A) $\operatorname{tg} 10^{\circ}+\operatorname{tg} 20^{\circ}$.
(B) $\operatorname{tg} 30^{\circ}$.
(C) $\frac{1}{2}\left(\operatorname{tg} 10^{\circ}+\operatorname{tg} 20^{\circ}\right)$.
(D) $\operatorname{tg} 15^{\circ}$.
(E) $\frac{1}{4} \operatorname{tg} 60^{\circ}$.
(35th American High School Mathematics Examination, 1984)
[Solution] From $\sin x+\sin y=2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}$, and $\cos x+\cos y=2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}$. $\therefore \quad \frac{\sin 10^{\circ}+\sin 20^{\circ}}{\cos 10^{\circ}+\cos 20^{\circ}}=\frac{\sin 15^{\circ}}{\cos 15^{\circ}}=\operatorname{tan} 15^{\circ}$. Therefore, the answer is $(D)$.