4. Simplify $\frac{\sin 7^{\circ}+\cos 15^{\circ} \cdot \sin 8^{\circ}}{\cos 7^{\circ}-\sin 15^{\circ} \cdot \sin 8^{\circ}}$ to get ( ).
A. $2+\sqrt{3}$
B. $\frac{2+\sqrt{3}}{2}$
C. $2-\sqrt{3}$
D. $\frac{2-\sqrt{3}}{2}$
4. C. Reason: By substituting $7^{\circ}$ with $15^{\circ}-8^{\circ}$, we have
$$
\begin{aligned}
\text { Original expression } & =\frac{\sin \left(15^{\circ}-8^{\circ}\right)+\cos 15^{\circ} \cdot \sin 8^{\circ}}{\cos \left(15^{\circ}-8^{\circ}\right)-\sin 15^{\circ} \cdot \sin 8^{\circ}}=\frac{\sin 15^{\circ} \cdot \cos 8^{\circ} \cdot \cos 15^{\circ} \cdot \sin 8^{\circ}+\cos 15^{\circ} \cdot \sin 8^{\circ}}{\cos 15^{\circ} \cdot \cos 8^{\circ}+\sin 15^{\circ} \cdot \sin 8^{\circ}-\sin 15^{\circ} \cdot \sin 8^{\circ}} \\
& =\tan 15^{\circ}=\tan \left(45^{\circ}-30^{\circ}\right)=\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1-\tan 45^{\circ} \cdot \tan 30^{\circ}}=2-\sqrt{3} .
\end{aligned}
$$