Hint: Use sine theorem. I'll post the complete answer when I get home.
I seem to be getting some problems with that $-1$ constant after $\arctan$. Are you sure those are the right formulas?
The idea to get angle $\beta$ in terms of $\alpha_1,\alpha_2$ is the following: Apply sine theorem in the two halves of triangle and obtain $\displaystyle \frac{\sin(\beta+\alpha_1+\alpha_2)}{\sin \beta}=\frac{\sin \alpha_2}{\sin\alpha_1}$. From here, using usual trigonometric formulas, you can extract $\tan \beta$, and then try and prove it has a closed form as presented in your formulas.
Here are some more details. We have $\frac{\sin \beta \cos(\alpha_1+\alpha_2)+\cos\beta\sin(\alpha_1+\alpha_2)}{\sin\beta}=\frac{\sin\alpha_2}{\sin\alpha_1}$. Then $\cos(\alpha_1+\alpha_2)+\cot\beta\sin(\alpha_1+\alpha_2)=\frac{\sin\alpha_2}{\sin\alpha_1}$. From here, you get a formula for $\cot \beta$, then you invert and get $\tan \beta$. I didn't manage to get your closed form, but if there is a way, this is the path. :)