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.
 A: 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
$$ \frac{\cos \alpha}{\cos \beta} = c = \frac{a}{b}$$
A: 
Sorry for bad drawing. The drawing is equal to the question.
$m(BAC)=\alpha$
$m(DBG)=\beta$
$x=a\tan \alpha$
$y=b\tan \beta$
Focus on EAD triangle.
$\tan (m(EAD))=\frac{x+y}{a+b}=\frac{a\tan \alpha+b\tan \beta}{a+b}=\frac{(a+b)\tan \frac{(\alpha+\beta)}{2}}{a+b}=\tan \frac{(\alpha+\beta)}{2}$
Thus 
$m(EAD)=m(HAC)=\frac{(\alpha+\beta)}{2}$
Let's calculate the angles: $m(BAD)$ and $m(ABD)$ and $m(ADB)$
$m(BAD)=m(BAC)-m(HAC)=\alpha-\frac{(\alpha+\beta)}{2}=\frac{(\alpha-\beta)}{2}$
$m(ABD)=90-\alpha+90+\beta=180-\alpha+\beta$
$m(ADB)=180-m(ABD)-m(BAD)=\frac{(\alpha-\beta)}{2}$
$m(ADB)=m(BAD)$
if so ABD triangle is an isosceles triangle 
Thus $AB=BD$
$AB=\frac{a}{\cos \alpha}$
$BD=\frac{b}{\cos \beta}$
$\frac{a}{\cos \alpha}=\frac{b}{\cos \beta}$
$\frac{\cos \alpha}{\cos \beta}=\frac{a}{b}$
A: Noting that
$$
\begin{align}
\tan\left(\frac{\alpha+\beta}{2}\right)
&=\frac{\sin(\alpha+\beta)}{1+\cos(\alpha+\beta)}\\
&=\frac{\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)}{1+\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)}\\
&=\frac{\tan(\alpha)+\tan(\beta)}{1+\sec(\alpha)\sec(\beta)-\tan(\alpha)\tan(\beta)}\tag{1}
\end{align}
$$
yields
$$
\begin{align}
\tan\left(\frac{\alpha+\beta}{2}\right)-\tan(\beta)
&=\frac{\tan(\alpha)+\tan(\beta)}{1+\sec(\alpha)\sec(\beta)-\tan(\alpha)\tan(\beta)}-\tan(\beta)\\
&=\frac{\tan(\alpha)-\sec(\alpha)\sec(\beta)\tan(\beta)+\tan(\alpha)\tan^2(\beta)}{1+\sec(\alpha)\sec(\beta)-\tan(\alpha)\tan(\beta)}\\
&=\frac{\tan(\alpha)\sec^2(\beta)-\sec(\alpha)\sec^2(\beta)\sin(\beta)}{1+\sec(\alpha)\sec(\beta)-\tan(\alpha)\tan(\beta)}\\
&=\frac{\tan(\alpha)-\sec(\alpha)\sin(\beta)}{\cos^2(\beta)+\sec(\alpha)\cos(\beta)-\tan(\alpha)\cos(\beta)\sin(\beta)}\\
&=\frac{\sec(\alpha)(\sin(\alpha)-\sin(\beta))}{\cos(\beta)\sec(\alpha)(\cos(\alpha)\cos(\beta)+1-\sin(\alpha)\sin(\beta))}\\
&=\frac{\sin(\alpha)-\sin(\beta)}{\cos(\beta)(\cos(\alpha+\beta)+1)}\tag{2}
\end{align}
$$
Symmetry and negation yields
$$
\tan(\alpha)-\tan\left(\frac{\alpha+\beta}{2}\right)=\frac{\sin(\alpha)-\sin(\beta)}{\cos(\alpha)(\cos(\alpha+\beta)+1)}\tag{3}
$$
Dividing $(2)$ by $(3)$ yields
$$
\frac{\tan\left(\frac{\alpha+\beta}{2}\right)-\tan(\beta)}{\tan(\alpha)-\tan\left(\frac{\alpha+\beta}{2}\right)}=\frac{\cos(\alpha)}{\cos(\beta)}\tag{4}
$$
Back to the original equation:
$$
a\tan(\alpha)+b\tan(\beta)=(a+b)\tan\left(\frac{\alpha+\beta}{2}\right)
$$
Dividing both sides by $b$:
$$
\frac{a}{b}\tan(\alpha)+\tan(\beta)=\left(\frac{a}{b}+1\right)\tan\left(\frac{\alpha+\beta}{2}\right)
$$
Collecting $\frac{a}{b}$ on the left:
$$
\frac{a}{b}\left(\tan(\alpha)-\tan\left(\frac{\alpha+\beta}{2}\right)\right)=\tan\left(\frac{\alpha+\beta}{2}\right)-\tan(\beta)
$$
Thus, by $(4)$,
$$
\frac{a}{b}=\frac{\tan\left(\frac{\alpha+\beta}{2}\right)-\tan(\beta)}{\tan(\alpha)-\tan\left(\frac{\alpha+\beta}{2}\right)}=\frac{\cos(\alpha)}{\cos(\beta)}
$$
as requested.
A: Did you try applying some identities?
$\tan\theta=\frac{\sin\theta}{\cos\theta}$ if $\cos\theta\neq0$.
$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$.
