Question

Let $\sin^{- 1}\frac{2}{\sqrt{13}} = \theta\&\cos^{- 1}\frac{3}{\sqrt{10}} = \phi$

$${sin\theta = \frac{2}{\sqrt{13}}\& cos\phi = \frac{3}{\sqrt{10}} }{tan(2\theta - 2\phi) = \frac{tan2\theta - tan2\phi}{1 + tan2\theta tan2\phi} }{\left( \because tan2\theta = \frac{2tan\theta}{1 - \tan^{2}\theta} \right) }{= \frac{\frac{12}{5} - \frac{3}{4}}{1 + \frac{12}{5} \cdot \frac{3}{4}} }{= \frac{33}{56}}$$