1. ## trigo problem

Given $\displaystyle \alpha + \beta = 60$degree
and $\displaystyle m = \frac {\sqrt 3 + tan \alpha}{\sqrt 3 + tan\beta}$

Prove that $\displaystyle tan\beta = \frac {2-m}{\sqrt 3 m}$

2. from

$\displaystyle m = \frac {\sqrt 3 + tan \alpha}{\sqrt 3 + tan\beta}$

you can obtain

$\displaystyle \tan{\beta} = \frac{\sqrt{3} + \tan{\alpha}-\sqrt{3}m}{m}$

now you have

$\displaystyle \frac{\sqrt{3} + \tan{\alpha}-\sqrt{3}m}{m} = \frac{2-m}{\sqrt{3}{m}}$

3. Originally Posted by bigwave
$\displaystyle \tan\left(\frac{\pi}{3}\right) = \sqrt{3}$

might help
$\displaystyle m = \frac {\sqrt 3 + tan (60-\beta)}{\sqrt 3 + tan \beta}$
$\displaystyle =\frac{\sqrt 3 + \frac {\sqrt 3 - tan \beta}{1+\sqrt3 tan \beta}}{\sqrt 3 + tan \beta}$
$\displaystyle = \frac {2 \sqrt 3 + 2 tan \beta}{\sqrt3 tan\beta (1+ \sqrt3 tan \beta)}$
???
don't know what to do next

4. Hello cakeboby
Originally Posted by cakeboby
Given $\displaystyle \alpha + \beta = 60$degree
and $\displaystyle m = \frac {\sqrt 3 + tan \alpha}{\sqrt 3 + tan\beta}$

Prove that $\displaystyle tan\beta = \frac {2-m}{\sqrt 3 m}$
$\displaystyle \alpha=60^o-\beta$

$\displaystyle \Rightarrow \tan\alpha = \frac{\sqrt3-\tan\beta}{1+\sqrt3\tan\beta}$ (1)

$\displaystyle \frac{2-m}{m}=\frac{2}{m}-1$
$\displaystyle =\frac{2(\sqrt3+\tan\beta)}{\sqrt3+\tan\alpha}-1$

$\displaystyle =\frac{2(\sqrt3+\tan\beta)}{\sqrt3+\left(\dfrac{\s qrt3-\tan\beta}{1+\sqrt3\tan\beta}\right)}-1$ from (1)
Simplify this. You'll find it comes to $\displaystyle \sqrt3\tan\beta$.