If $\sin\alpha+\sin\beta=a$ and $\cos\alpha-\cos\beta=b$, then what is $\tan(\frac{\alpha-\beta}{2})$? If I square both the equations 
$$2+2\sin(\alpha-\beta)=a^2+b^2$$
$$\sin(\alpha-\beta)=\frac{a^2+b^2-2}{2}$$
Since $\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$, then $$\sin(\alpha-\beta)=\frac{2\tan\frac{\alpha-\beta}{2}}{1+\left(\tan\frac{\alpha-\beta}{2}\right)^2}$$
It’s obviously the too long to solve, so is there a shorter way to do this?

Since $\frac{a}{2}=\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$ while $-\frac{b}{2}=\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$,  $\tan\frac{\alpha-\beta}{2}=-\frac{b}{a}$.
