$ \begin{aligned} & 2(\cos \beta-\cos \alpha)+\cos \alpha \cdot \cos \beta=1 \\ & 2 \cos \beta+\cos \alpha \cos \beta-2 \cos \alpha=1 \\ & 2 \cos \beta(2+\cos \alpha)=1+2 \cos \alpha \end{aligned} $ $ \cos \beta=\frac{1+2 \cos \alpha}{2+\cos \alpha}=\frac{1+2\left(\frac{1-\tan ^2 \frac{\alpha}{2}}{1+\tan ^2 \frac{\alpha}{3}}\right)}{2+\frac{1-\tan ^2 \frac{\alpha}{2}}{1+\tan ^2 \frac{\alpha}{2}}} $ $\begin{aligned} & \frac{1-\tan ^2 \frac{\beta}{2}}{1+\tan ^2 \frac{\beta}{2}}=\frac{1+\tan ^2 \frac{\alpha}{2}+2-2 \tan ^2 \frac{\alpha}{2}}{2+2 \tan ^2 \frac{\alpha}{2}+1-\tan ^2 \frac{\alpha}{2}} \\ & \frac{1-\tan ^2 \frac{\beta}{2}}{1+\tan ^2 \frac{\beta}{2}}=\frac{3-\tan ^2 \frac{\alpha}{2}}{3+\tan ^2 \frac{\alpha}{2}} \\ & \frac{1-\tan ^2 \frac{\beta}{2}}{1+\tan ^2 \frac{\beta}{2}}=\frac{1-\frac{1}{3} \tan ^2 \frac{\alpha}{2}}{1+\frac{1}{3} \tan ^3 \frac{\alpha}{2}} \\ & \frac{1}{3} \tan ^2 \frac{\alpha}{2}=\tan ^2 \frac{\beta}{2} \\ & \tan ^2 \frac{\alpha}{2}=3 \tan 2 \frac{\beta}{2} \\ & \left(\tan \frac{\alpha}{2}-\sqrt{3} \tan \frac{\beta}{2}\right)\left(\tan \frac{\alpha}{2}+\sqrt{3} \tan \frac{\beta}{2}\right)=0\end{aligned}$