Question

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

Answer 1

$\frac{m}{n}=\frac{\tan (\theta +{120}^o)}{\tan (\theta -{30}^o)}=\frac{\tan A}{\tan B}=\frac{\sin A\cos B}{\cos A\sin B}$
where $A=\theta +{120}^o;B=\theta -{30}^o$
Applying componendo and dividendo, we get
$\frac{m+n}{m-n}=\frac{\sin (A+B)}{\sin (A-B)}=\frac{\sin (2\theta +{90}^o)}{\sin {150}^o}=\frac{\cos 2\theta }{\sin {30}^o}=2\cos 2\theta$
$\therefore \cos 2\theta =\frac{m+n}{2(m-n)}$