Another trig prob

Ans) \frac{m}{n} = \frac{tan\ (\theta +120)}{tan(\theta -30)} = \frac{cos(\theta +30)cos(\theta -30)}{sin(\theta +30)sin(\theta -30)}

Therefore, \frac{m}{n} = \frac{cos^{2}\ 30 - sin^{2}\theta }{sin^{2}\ 30 - sin^{2}\theta} = \frac{3/4 -sin^{2}\theta }{1/4 - sin^{2}\theta}

On applying componendo and dividendo, we get

\frac{m+n}{m-n} = \frac{1-2\ sin^{2}\theta }{(1/2)} = 2\ cos\ 2\theta

Therefore, (b) option