A) \[\frac{m+n}{m-n}\]
B) \[\frac{m-n}{m+n}\]
C) \[\frac{m-n}{2(m+n)}\]
D) \[\frac{m+n}{2(m-n)}\]
E) \[\frac{2(m+n)}{m-n}\]
Correct Answer: D
Solution :
\[m\tan (\theta -30{}^\circ )=n\tan (\theta +120{}^\circ )\] \[\Rightarrow m\left( \frac{\tan \theta -\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}\tan \theta } \right)=-\,n\left( \frac{-\tan \theta +\sqrt{3}}{1+\sqrt{3}\tan \theta } \right)\] \[\Rightarrow \]\[m[{{(\sqrt{3}\tan \theta )}^{2}}-1]=-n(-{{\tan }^{2}}\theta +3)\] \[\Rightarrow \] \[3m{{\tan }^{2}}\theta -m=n{{\tan }^{2}}\theta -3n\] \[\Rightarrow \] \[{{\tan }^{2}}\theta =\frac{m-3n}{3m-n}\] Now, \[\cos 2\theta =\frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }\] \[=\frac{1-\frac{m-3n}{3m-n}}{1+\frac{m-3n}{3m-n}}=\frac{3m-n-m+3n}{3m-n+m-3n}\] \[=\frac{2(m+n)}{4(m-n)}=\frac{m+n}{2(m-n)}\]
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