How to prove that $\frac{\cos\alpha}{\cos\beta}=a/b$ If $\alpha \not= \beta$, and $$ a\tan \alpha+b\tan\beta=(a+b)\tan\frac{\alpha+\beta}{2}$$ then can we prove that $\frac{\cos\alpha}{\cos\beta}=\frac{a}{b}$?
Seems like I am stuck on this one.

This is true if we assume $\displaystyle \sin\left(\frac{\alpha -\beta}{2}\right) \neq 0$ and $\displaystyle b \neq 0$.
Let $\displaystyle \frac{a}{b} = c$
Rewrite your equation as
$$c\tan \alpha + \tan \beta = (c+1)\tan(\frac{\alpha+\beta}{2})$$
$$ c\left(\tan \alpha - \tan\left(\frac{\alpha+\beta}{2}\right)\right) = \tan\left(\frac{\alpha+\beta}{2}\right) - \tan \beta$$
Using $\displaystyle \tan x + \tan y = \frac{\sin(x+y)}{\cos x \cos y}$ we get
$$ c \cdot \frac{\sin\left(\frac{\alpha-\beta}{2}\right)}{\cos \alpha \cos \left(\frac{\alpha+\beta}{2}\right)} =\frac{\sin\left(\frac{\alpha-\beta}{2}\right)}{\cos \beta \cos \left(\frac{\alpha+\beta}{2}\right)}$$
and so
$$ \frac{\cos \alpha}{\cos \beta} = c = \frac{a}{b}$$
