2015/117) If $m\tan(a-30^\circ)=n\tan(a+120^\circ)$ show that $\cos2a=\frac{m+n}{2(m-n)}$

We have $\tan (a+ 120^\circ) = - \cot(a + 30^\circ)$ using $\tan (x+90^\circ) = - \cot x$
so $m\tan(a-30^\circ)= -n \cot ( a+ 30^\circ)$
so $\tan (a+30^\circ) \tan (a-30^\circ) = \dfrac{-n}{m}$
or $\dfrac{\tan a + \tan 30^\circ}{1- \tan a \tan 30^\circ} * \dfrac{\tan a -\tan 30^\circ}{1 + \tan a \tan 30^\circ)} = \dfrac{-n}{m}$
or $\dfrac{tan ^2 a - tan ^2 30^\circ}{1- tan ^2 a tan ^2 30^\circ} = \dfrac{n}{m}$
or $\dfrac{tan ^2 a - \frac{1}{3}}{1- tan ^2 a \frac{1}{3}} = \dfrac{n}{m}$
or $\dfrac{3 tan ^2 a -1}{3 - tan ^2 a} = \dfrac{n}{m}$
or $\dfrac{1-3 tan ^2 a}{3- tan ^2 a} = \dfrac{n}{m}$
use componendo dividendo to get
$\dfrac{4 - 4 \tan ^2 a}{- 2 - 2 \tan ^2 a} = \dfrac{n+m}{n-m}$
or $2\dfrac{1 - tan ^2 a}{1 + 2 tan ^2 a} = \dfrac{n+m}{m-n}$
or $\dfrac{1 - tan ^2 a}{1 + 2 tan ^2 a} = \dfrac{n+m}{2(m-n)}$
or $\cos 2a = \dfrac{n+m}{2(m-n)}$