$\mathrm{xtan}\left(\mathrm{\theta }-\frac{\mathrm{\pi }}{6}\right)=\mathrm{ytan}\left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)$then  $\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}-\mathrm{y}}=$

$$\begin{gathered} \mathrm{xtan}\left(\mathrm{\theta }-\frac{\mathrm{\pi }}{6}\right)=\mathrm{ytan}\left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right) \\ \Rightarrow \frac{\mathrm{x}}{\mathrm{y}}=\frac{\tan \left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)}{\tan \left(\mathrm{\theta }-\frac{\mathrm{\pi }}{6}\right)} \end{gathered}$$
By componendo and dividendo  $\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}-\mathrm{y}}=\frac{\tan \left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)+\tan \left(\mathrm{\theta }-\frac{\mathrm{\pi }}{6}\right)}{\tan \left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)-\tan \left(\mathrm{\theta }-\frac{\mathrm{\pi }}{6}\right)}$
$\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}-\mathrm{y}}=\frac{\sin \left(2\mathrm{\theta }+\frac{\mathrm{\pi }}{2}\right)}{\sin \left(\frac{5\mathrm{\pi }}{6}\right)}=2\cos 2\mathrm{\theta }$