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?
 A: $$\sin\alpha + \sin\beta = 2\sin(\frac{\alpha+\beta}{2})\cos(\frac{\alpha-\beta}{2}).$$
$$\cos\alpha + \cos\beta = 2\cos(\frac{\alpha+\beta}{2})\cos(\frac{\alpha-\beta}{2}).$$
So, you get the conclusion.
A: 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:
$$
=\dfrac{a+b}{1-ab}=\dfrac{\tan \alpha/2+\tan \beta/2}{1-\tan \alpha/2 \tan \beta/2}= \tan \left(\dfrac{\alpha +\beta}{2} \right)
$$
A: Assume that $$P=(\cos\ a,\sin\ b),\ Q=(\cos\ b,\sin\ b)$$ Then what is
slope of line passing through $-Q,\ P$ :
For convenience $0<a<b<\pi/2$, let $R=(0,1)$. Then $$\angle\
ROQ=\frac{\pi}{2}-b,\ O=(0,0)$$
and $$\angle\ Q(-Q)P = \frac{b-a}{2} $$
When $\angle\ PP'(-Q) = \pi/2$, then $\angle\ P(-Q)P'=
\frac{a+b}{2}$
Hence $$ \tan\ \frac{a+b}{2} = \frac{{\rm difference \ of}\ y-{\rm
coordinates\ of}\ P,\ -Q }{{\rm difference \ of}\ x-{\rm coordinates
\ of}\ P,\ -Q } $$
