<s>[INST] 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?
 [/INST] 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 $$ \tan\ \frac{a+b}{2} = \frac{{\rm difference \ of}\ y-{\rm
coordinates\ of}\ P,\ -Q }{{\rm difference \ of}\ x-{\rm coordinates
\ of}\ P,\ -Q } $$
</s>