# 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.

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This is not true. $\alpha = 0$, $\beta = 2\pi$. Are there any restrictions? Like maybe they are acute? – Aryabhata Feb 29 '12 at 6:35
I wonder if it might help to recall that $\tan((\alpha+\beta)/2)=(\sin\alpha+\sin\beta)/(\cos\alpha+\cos\beta)$? – Michael Hardy Feb 29 '12 at 15:12

This is true if we assume $\displaystyle \sin\left(\frac{\alpha -\beta}{2}\right) \neq 0$ and $\displaystyle b \neq 0$.

Let $\displaystyle \frac{a}{b} = c$

$$c\tan \alpha + \tan \beta = (c+1)\tan(\frac{\alpha+\beta}{2})$$

$$c\left(\tan \alpha - \tan\left(\frac{\alpha+\beta}{2}\right)\right) = \tan\left(\frac{\alpha+\beta}{2}\right) - \tan \beta$$

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

$$\frac{\cos \alpha}{\cos \beta} = c = \frac{a}{b}$$

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$\displaystyle\tan(x)+\tan(y)=\frac{\sin(x+y)}{\cos(x)\cos(y)}$ makes this a lot simpler. (+1) – robjohn Feb 29 '12 at 13:09

Sorry for bad drawing. The drawing is equal to the question.

$m(BAC)=\alpha$

$m(DBG)=\beta$

$x=a\tan \alpha$

$y=b\tan \beta$

$\tan (m(EAD))=\frac{x+y}{a+b}=\frac{a\tan \alpha+b\tan \beta}{a+b}=\frac{(a+b)\tan \frac{(\alpha+\beta)}{2}}{a+b}=\tan \frac{(\alpha+\beta)}{2}$

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 $AB=BD$

$AB=\frac{a}{\cos \alpha}$

$BD=\frac{b}{\cos \beta}$

$\frac{a}{\cos \alpha}=\frac{b}{\cos \beta}$

$\frac{\cos \alpha}{\cos \beta}=\frac{a}{b}$

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I'd been looking for the picture proof, and you beat me to it! :) But ... Isn't the similarity stuff irrelevant? The key is that $\triangle ABD$ is isosceles (say, with leg $r$), so that $a = r\cos\alpha$ and $b = r\cos\beta$ ... and we're done, right? – Blue Feb 29 '12 at 14:24
(Missed the comment-editing time window.) That $\angle HAC = (\alpha+\beta)/2$ simply restates the assumed relation, since $(a+b)\tan(\angle HAC) = x+y = a\tan\alpha+b\tan\beta$. There's no need to derive this angle's measure; we're given it. – Blue Feb 29 '12 at 14:33
If We do not have the relation $a\tan \alpha+b\tan \beta=(a+b)\tan \frac{(\alpha+\beta)}{2}$ in question, We cannot say $m(HAC)=\frac{(\alpha+\beta)}{2}$ – Mathlover Feb 29 '12 at 14:54
Right, and you used the relation in your computation of $k$. I'm just saying that the relation is immediately observable in your clever diagram: In $\triangle DAE$, clearly $\tan\angle DAE = (x+y)/(a+b)$. We're told that $(\alpha+\beta)/2$ matches this tangent value, so there's nothing elaborate to prove here. Thales is certainly overkill. – Blue Feb 29 '12 at 15:46
I saw what you mean. you are correct.Blue triangle is enough to proof. Thanks in advice. – Mathlover Feb 29 '12 at 19:28

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.

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Did you try applying some identities? $\tan\theta=\frac{\sin\theta}{\cos\theta}$ if $\cos\theta\neq0$. $\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$.

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