Prove that $m\tan (\theta-30°)=n\tan (\theta+120°)$ If $m\tan (\theta-30°)=n\tan (\theta+120°)$ then prove that :
$$\cos 2\theta=\frac{m+n}{2(m-n)}$$
My attempt\
Here,
$$m\tan (\theta-30°)=n\tan (\theta+120)$$
$$\frac{\tan (\theta-30°)}{\tan (\theta+120°)}=\frac{n}{m}$$.
Now, what should I do next?

$$m\tan (\theta-30^\circ)=n\tan (\theta+120^\circ)$$
$$\frac{\tan (\theta+120^\circ)}{\tan (\theta-30^\circ)}=\frac{m}{n}$$
Applying Componendo-Dividendo, we get
$$\frac{\tan (\theta+120^\circ)+\tan (\theta-30^\circ)}{\tan (\theta+120^\circ)-\tan (\theta-30^\circ)}=\frac{m+n}{m-n}$$
Now expand the L.H.S., keeping in mind that $\tan 30^\circ=\frac{1}{\sqrt3}$ and $\tan 120^\circ=-\sqrt3$
As Lab suggested, you can go as follows:
$$\frac{\frac{\sin (\theta+120^\circ)}{\cos (\theta+120^\circ)}+\frac{\sin (\theta-30^\circ)}{\cos (\theta-30^\circ)}}{\frac{\sin (\theta+120^\circ)}{\cos (\theta+120^\circ)}-\frac{\sin (\theta-30^\circ)}{\cos (\theta-30^\circ)}}=\frac{m+n}{m-n}$$
Then you have, $$\frac{\frac{\sin (\theta+120^\circ)\cos (\theta-30^\circ)+\cos (\theta+120^\circ)\sin (\theta-30^\circ)}{\cos (\theta+120^\circ)\cos (\theta-30^\circ)}}{\frac{\sin (\theta+120^\circ)\cos (\theta-30^\circ)-\cos (\theta+120^\circ)\sin (\theta-30^\circ)}{\cos (\theta+120^\circ)\cos (\theta-30^\circ)}}=\frac{m+n}{m-n}$$
The denominators cancel out to again give you
$$\frac{\sin (\theta+120^\circ)\cos (\theta-30^\circ)+\cos (\theta+120^\circ)\sin (\theta-30^\circ)}{\sin (\theta+120^\circ)\cos (\theta-30^\circ)-\cos (\theta+120^\circ)\sin (\theta-30^\circ)}=\frac{m+n}{m-n}$$
$$\frac{\sin (2\theta + 90^\circ)}{\sin 150^\circ}=\frac{m+n}{m-n}$$
Can you complete the rest now?
Hope this helps.
