3 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
A. 1
B. $\frac{1}{2}$
C. $\sqrt{2}$
D. $\frac{\sqrt{2}}{2}$
3 A. From the problem, we have
$$
\frac{\cos \left(\alpha+\frac{\beta}{2}\right)+\cos \left(\alpha-\frac{\beta}{2}\right)}{\cos \left(\alpha-\frac{\beta}{2}\right)}+\frac{\cos \left(\beta+\frac{\alpha}{2}\right)+\cos \left(\beta-\frac{\alpha}{2}\right)}{\cos \left(\beta-\frac{\alpha}{2}\right)}=2,
$$

which means
$$
\frac{\cos \left(\alpha+\frac{\beta}{2}\right)}{\cos \left(\alpha-\frac{\beta}{2}\right)}=-\frac{\cos \left(\beta+\frac{\alpha}{2}\right)}{\cos \left(\beta-\frac{\alpha}{2}\right)} .
$$

By the theorem of the ratio of sums and differences, 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
$$
\frac{\cos \alpha}{2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}}=\frac{2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}}{\cos \beta} .
$$

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

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