Double $\frac{\cos \alpha \cos \frac{\beta}{2}}{\cos \left(a-\frac{\beta}{2}\right)}+\frac{\cos \beta \cos \frac{\alpha}{2}}{\cos \left(\beta-\frac{\alpha}{2}\right)}-1$. 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 
\[
\frac{\cos \left(\alpha+\frac{\beta}{2}\right)+\cos \left(\alpha-\frac{\beta}{2}\right)}{\cos \left(\alpha-\frac{\beta}{2}\right)}+\frac{\cos \left(\beta+\frac{\alpha}{2}\right)+\cos \left(\beta-\frac{\alpha}{2}\right)}{\cos \left(\beta-\frac{\alpha}{2}\right)}=2,
\]
that is,

By the theorem of the ratio of sums and differences, we get
that is,
\[
\begin{array}{l}
\text { that is } \frac{\cos \alpha}{2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}}=\frac{2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}}{\cos \beta} . \\
\text { Hence } \cos \alpha \cos \beta=4 \sin ^{2} \frac{\alpha}{2} \sin ^{2} \frac{\beta}{2}=(1-\cos \alpha)(1-\cos \beta) . \\
\text { Hence } \\
\cos \alpha+\cos \beta=1 .
\end{array}
\]