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?

Another approach:
Put: $\tan (\alpha/2)=a$ and $ \tan (\beta/2)=b$.
Than:
$$
\sin \alpha= \dfrac{2a}{1+a^2} \qquad \cos \alpha=\dfrac{1-a^2}{1+a^2}
$$
$$
\sin \beta= \dfrac{2b}{1+b^2} \qquad \cos \beta=\dfrac{1-b^2}{1+b^2}
$$
and:
$$
\frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} = \dfrac{2a(1+b^2)+2b(1+a^2)}{(1-a^2)(1+b^2)+(1-b^2)(1+a^2)}
$$
that, after a bit of algebra, becomes:
$$
=\dfrac{a+b}{1-ab}=\dfrac{\tan \alpha/2+\tan \beta/2}{1-\tan \alpha/2 \tan \beta/2}= \tan \left(\dfrac{\alpha +\beta}{2} \right)
$$
