Let $ \sin \theta = \frac{2}{\sqrt{13}} \Rightarrow \theta $, and $ \cos \phi = \frac{3}{\sqrt{10}} \Rightarrow \phi $

$ \sin \theta = \frac{2}{\sqrt{13}},; \cos \phi = \frac{3}{\sqrt{10}} $

$ \tan(2\theta - 2\phi) = \frac{\tan 2\theta - \tan 2\phi}{1 + \tan 2\theta \tan 2\phi} $

$ \tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta} $

$ = \frac{12}{5} $

$ \tan 2\phi = \frac{3}{4} $

$ \Rightarrow \frac{\frac{12}{5} - \frac{3}{4}}{1 + \frac{12}{5} \cdot \frac{3}{4}} = \frac{33}{56} $