Compound angle Vs Double angle Vs Componendo and Dividendo

2. Prove that  $ \displaystyle \cos 2\theta =\frac{{1+{{{\tan }}^{2}}\theta }}{{1+{{{\tan }}^{2}}\theta }}$ . Hence prove that $ \displaystyle \cos 2\theta =\frac{{m+n}}{{2(m-n)}}$ given that $ \displaystyle m\tan (\theta -\frac{\pi }{6})=n\tan (\theta +\frac{{2\pi }}{3})$ .

$ \displaystyle \begin{array}{l}\cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta \\\ \ \ \ \ \ \ \ \ \ =\frac{{{{{\cos }}^{2}}\theta -{{{\sin }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta +{{{\sin }}^{2}}\theta }}\\\ \ \ \ \ \ \ \ \ \ =\frac{{1-\frac{{{{{\sin }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta }}}}{{1+\frac{{{{{\sin }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta }}}}\\\ \ \ \ \ \ \ \ \ \ =\frac{{1-{{{\tan }}^{2}}\theta }}{{1+{{{\tan }}^{2}}\theta }}\\m\tan (\theta -\frac{\pi }{6})=n\tan (\theta +\frac{{2\pi }}{3})\\m(\frac{{\tan \theta -\tan {{{30}}^{\circ }}}}{{1+\tan \theta \tan {{{30}}^{\circ }}}})=n(\frac{{\tan \theta +\tan {{{120}}^{\circ }}}}{{1-\tan \theta \tan {{{120}}^{\circ }}}})\\m(\frac{{\tan \theta -\frac{1}{{\sqrt{3}}}}}{{1+\frac{1}{{\sqrt{3}}}\tan \theta }})=n(\frac{{\tan \theta -\sqrt{3}}}{{1+\sqrt{3}\tan \theta }})\\m(\frac{{\sqrt{3}\tan \theta -1}}{{\sqrt{3}+\tan \theta }})=n(\frac{{\tan \theta -\sqrt{3}}}{{1+\sqrt{3}\tan \theta }})\\\frac{m}{n}=\frac{{\tan \theta -\sqrt{3}}}{{1+\sqrt{3}\tan \theta }}\times \frac{{\sqrt{3}+\tan \theta }}{{\sqrt{3}\tan \theta -1}}\\\frac{m}{n}=\frac{{{{{\tan }}^{2}}\theta -3}}{{3{{{\tan }}^{2}}\theta -1}}\\\frac{{m+n}}{{m-n}}=\frac{{4{{{\tan }}^{2}}\theta -4}}{{-2{{{\tan }}^{2}}\theta -2}}\left[ {By\ Componendo\ and\ Dividendo} \right]\\\frac{{m+n}}{{m-n}}=\frac{{2(1-{{{\tan }}^{2}}\theta )}}{{1+{{{\tan }}^{2}}\theta }}\\\frac{{m+n}}{{2(m-n)}}=\frac{{1-{{{\tan }}^{2}}\theta }}{{1+{{{\tan }}^{2}}\theta }}\\\therefore \ \ \cos 2\theta =\frac{{m+n}}{{2(m-n)}}\end{array}$