Trig3.9

If $m\tan (\theta-30)=n\tan (\theta+120)\,$ show that $\cos 2\theta=\frac{m+n}{2(m-n)}\,$

$m[\frac{\tan\theta-\tan 30}{1+\tan \theta \tan 30}]=n[\frac{\tan \theta+\tan 120}{1-\tan \theta \tan 120}]\,$

$m[\frac{\tan \theta-\frac{1}{\sqrt{3}}}{1+\tan \theta \frac{1}{\sqrt{3}}}]=n[\frac{\tan \theta-\sqrt{3}}{1+\tan \theta \sqrt{3}}]\,$

Cross multiplying,we get

$m(\sqrt{3} \tan \theta-1)(\sqrt{3} \tan \theta+1)=n(\sqrt{3}+\tan \theta)(\tan \theta-\sqrt{3}\,$

$m(3\tan^2 \theta-1)=n(\tan^2 \theta-3)\,$

$\tan^2 \theta=\frac{m-3n}{m+n}\,$

Therefore, $\cos 2\theta=\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\,$

$\frac{1-\frac{m-3n}{3m-n}}{1+\frac{m-3n}{3m-n}}\,$

$\cos 2\theta=\frac{m+n}{2(m-n)}\,$