8 Given $\frac{\cos \alpha \cos \frac{\beta}{2}}{\cos \left(\alpha-\frac{\beta}{2}\right)}+\frac{\cos \beta \cos \frac{\alpha}{2}}{\cos \left(\beta-\frac{\alpha}{2}\right)}=1$, then the value of $\cos \alpha+\cos \beta$ is . $\qquad$
(8) 1 Hint: From the problem, we have
$$
\begin{array}{l}
\frac{\cos \left(\alpha+\frac{\beta}{2}\right)+\cos \left(\alpha-\frac{\beta}{2}\right)}{\cos \left(\alpha-\frac{\beta}{2}\right)}+\frac{\cos \left(\beta+\frac{\alpha}{2}\right)+\cos \left(\beta-\frac{\alpha}{2}\right)}{\cos \left(\beta-\frac{\alpha}{2}\right)}=2 . \\
\text { That is } \frac{\cos \left(\alpha+\frac{\beta}{2}\right)}{\cos \left(\alpha-\frac{\beta}{2}\right)}=-\frac{\cos \left(\beta+\frac{\alpha}{2}\right)}{\cos \left(\beta-\frac{\alpha}{2}\right)} \text {. } \\
\end{array}
$$

By the theorem of proportion, we get
$$
\frac{\cos \alpha \cos \frac{\beta}{2}}{\sin \alpha \sin \frac{\beta}{2}}=\frac{\sin \beta \sin \frac{\alpha}{2}}{\cos \beta \cos \frac{\alpha}{2}},
$$

which means
$$
\frac{\cos \alpha}{2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}}=\frac{2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}}{\cos \alpha},
$$

thus
$$
\cos \alpha \cos \beta=4 \sin ^{2} \frac{\alpha}{2} \sin ^{2} \frac{\beta}{2}=(1-\cos \alpha)(1-\cos \beta) .
$$

Therefore, $\cos \alpha+\cos \beta=1$.