# Thread: trigo problem

1. ## trigo problem

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

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

2. from

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

you can obtain

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

now you have

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

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

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

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

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

$\alpha=60^o-\beta$

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

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

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