trigonometry - Proving that $frac{sin alpha + sin beta}{cos alpha + cos beta} =tan left ( frac{alpha+beta}{2} right )$

Using double angle identities a total of four times, one for each expression in the left hand side, I acquired this.

$$\frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} = \frac{\sin \left ( \frac{\alpha}{2}\right ) \cos \left ( \frac{\alpha}{2}\right ) + \sin \left ( \frac{\beta}{2}\right ) \cos \left ( \frac{\beta}{2}\right )}{\cos^2 \left ( \frac{\alpha}{2} \right) - \sin ^2 \left ( \frac{\beta}{2} \right )}$$

But I know that if $\alpha$ and $\beta$ are angles in a triangle, then this expression should simplify to

$$\tan \left ( \frac{\alpha + \beta}{2} \right )$$

I can see that the denominator becomes $$\cos \left ( \frac{\alpha + \beta}{2} \right )$$

But I cannot see how the numerator becomes

$$\sin \left ( \frac{\alpha + \beta}{2} \right )$$

What have I done wrong here?