2015/051)If $m\tan(a-30^\circ)=n\tan(a+120^\circ)$ show that $\cos2a=\dfrac{m+n}{2(m-n)}$

we have $\tan (a+ 120^\circ) = - \cot(a + 30^\circ)$ using $\tan (x+90^\circ) = - cot\, x$
so $mtan(a-30^\circ)= -n \cot ( a+ 30^\circ)$
so $\tan (a+30^\circ) tan (a-30^\circ) = - \dfrac{n}{m}$
$\dfrac{\tan\, a + \tan\, 30^\circ}{1- \tan\, a \tan\, 30^\circ}\dfrac{tan\, a -\tan\, 30^\circ}{1 + \tan\, a \tan\, 30^\circ} = -\dfrac{n}{m}$
or $\dfrac{\tan ^2 a - \tan ^2 30^\circ}{1- \tan ^2 a \tan ^2 30^\circ} = -\dfrac{n}{m}$

or $\dfrac{\tan ^2 30^\circ-\tan ^2 a}{1- \tan ^2 a \tan ^2 30^\circ} = \dfrac{n}{m}$

or $\dfrac{\frac{1}{3}-\tan ^2 a}{1- \frac{1}{3}\tan ^2 a} = \dfrac{n}{m}$

 or $\dfrac{1-3 \tan ^2 a}{3- \tan ^2 a} = \dfrac{n}{m}$

  using componendo dividendo to get

$\dfrac{4 - 4 \tan ^2 a}{- 2 - 2 \tan ^2 a} = \dfrac{n+m}{n-m}$

or 

$2 \dfrac{1 - 1 \tan ^2 a}{1 + \tan ^2 a} = \dfrac{n+m}{m-n}$

or 

$2 \dfrac{1 - \tan ^2 a}{sec ^2 a} = \dfrac{n+m}{m-n}$

or

$2 (1 - \tan ^2 a)(cos ^2 a) = \dfrac{n+m}{m-n}$

 or $(\cos^2 a-\sin ^2 a) = \dfrac{m+n}{2(m-n)}$
or $\cos 2a = \dfrac{m+n}{2(m-n)}$

This is solved at https://uk.answers.yahoo.com/question/index?qid=20120601212526AAguVnm