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}
$
Thanks in advance.
$\displaystyle \alpha=60^o-\beta$
$\displaystyle \Rightarrow \tan\alpha = \frac{\sqrt3-\tan\beta}{1+\sqrt3\tan\beta}$ (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 $\displaystyle \sqrt3\tan\beta$.
Grandad