Question

If $m\tan (\theta -{30}^{\circ })=n\tan (\theta +{120}^{\circ })$, then prove that $\cos 2\theta =\frac{m+n}{2(m-n)}$.

Answer 1

$m\tan (\theta -30)=n\tan (\theta +120)\Rightarrow \frac{m}{n}=\frac{\tan (\theta +120)}{\tan (\theta -30)}$
Applying Componendo and dividendo
$\Rightarrow \frac{m+n}{m-n}=\frac{\tan (\theta +120)+\tan (\theta -30)}{\tan (\theta +120)-\tan (\theta -30)}$
$\Rightarrow \frac{m+n}{m-n}=\frac{\frac{\sin (\theta +120)}{\cos (\theta +120)}+\frac{\sin (\theta -30)}{\cos (\theta -30)}}{\frac{\sin (\theta +120)}{\cos (\theta +120)}-\frac{\sin (\theta -30)}{\cos (\theta -30)}}$
$\Rightarrow \frac{m+n}{m-n}=\frac{\sin (\theta +120)\cos (\theta -30)+\sin (\theta -30)\cos (\theta +120)}{\sin (\theta +120)\cos (\theta -30)-\sin (\theta -30)\cos (\theta +120)}$
$\Rightarrow \frac{m+n}{m-n}=\frac{\sin (\theta +120+\theta -30)}{\sin (\theta +120-\theta +30)}=\frac{\sin (2\theta +90)}{\sin 180}=2\cos 2\theta$
$\Rightarrow \cos 2\theta =\frac{1}{2}\left(\frac{m+n}{m-n}\right)$