# Solving for $\beta$ in $\frac {m\tan(\alpha-\beta)} {\cos^2\beta} = \frac {n\tan\beta}{\cos^2(\alpha-\beta)}$ [closed]

It is given that $$\frac {m\tan(\alpha-\beta)} {\cos^2\beta} = \frac {n\tan\beta}{\cos^2(\alpha-\beta)}$$

Show that: $$\beta = \frac{1}{2}\left(\alpha - \arctan\frac{(n-m)\tan\alpha}{n+m}\right)$$

I tried solving the given condition by changing tangent into the ratio of sine and cosine. I reached the following conclusion:

$$m\sin(2\alpha-2\beta)-n\sin(2\beta) = 0$$

I couldn't factor it out any further. Can anyone help me out?

## closed as off-topic by Namaste, Brahadeesh, user91500, Vladhagen, NosratiSep 17 '18 at 16:09

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• But the $$\beta$$ is also on the right-hand side of your solution! – Dr. Sonnhard Graubner Sep 16 '18 at 17:09
• Yes, there's a typo: the RHS should have $\tan \alpha$. The solution below indicates that. – Andreas Sep 16 '18 at 21:16

$\dfrac{n-m}{n+m}$ demands Componendo and Dividendo

We have $$\dfrac nm=\dfrac{\sin2(\alpha-\beta)}{\sin2\beta}\text{ using }\sin2A=2\sin A\cos A$$

Applying Componendo and Dividendo,

$$\dfrac{n-m}{n+m}=\dfrac{\sin2(\alpha-\beta)-\sin2\beta}{\sin2(\alpha-\beta)+\sin2\beta}$$

Apply Prosthaphaeresis Formulas and simplify

Let $\beta = \frac12 (\alpha - \gamma$). Then, with your intermediate step:

$m\sin(2\alpha-2\beta)-n\sin(2\beta) = 0$ it follows

$m\sin(\alpha + \gamma)-n\sin(\alpha - \gamma) = 0$

With the usual trig theorem for the sum of two angles this is

$m(\sin(\alpha) \cos(\gamma) + \cos(\alpha) \sin(\gamma)) -n(\sin(\alpha) \cos(\gamma) - \cos(\alpha) \sin(\gamma))= 0$

or

$(m-n)\sin(\alpha) \cos(\gamma) + (m+n) \cos(\alpha) \sin(\gamma)= 0$

or

$(n-m)\tan(\alpha) = (m+n) \tan(\gamma)$

hence

$\gamma = \arctan \left[ \frac{(n-m)\tan(\alpha)}{m+n} \right]$

and this is the required answer. Note there should be an $\alpha$ on the RHS - see Dr. Sonnhard Graubner'S comment!

I think you made a typo there. It should be $$\beta=\frac{1}{2}\left(\alpha-\arctan\frac{(n-m)\tan\alpha}{n-m}\right)$$ instead of $$\beta=\frac{1}{2}\left(\alpha-\arctan\frac{(n-m)\tan\beta}{n-m}\right)$$

Back to solution We want to show $$\beta=\frac{1}{2}\left(\alpha-\arctan\frac{(n-m)\tan\alpha}{n-m}\right)$$ i.e., we want: $$\tan(\alpha-2\beta) = \frac{n-m}{n+m}\tan\alpha$$

But we know $$\frac{m}{n}=\frac{\sin 2\beta}{\sin(2\alpha-2\beta)}$$

So we want to show $$\tan(\alpha-2\beta)=\frac{\sin(2\alpha-2\beta)-\sin2\beta}{\sin(2\alpha-2\beta)+\sin 2\beta}\tan\alpha$$ which is immediate since $$\frac{\sin(2\alpha-2\beta)-\sin2\beta}{\sin(2\alpha-2\beta)+\sin 2\beta} = \frac{2\sin(\alpha-2\beta)\cos\alpha}{2\sin\alpha\cos(\alpha-2\beta)}=\frac{\tan(\alpha-2\beta)}{\tan\alpha}$$