Question

If $\cos (\alpha +\beta )=\frac{3}{5}$ and $\sin (\alpha -\beta )=\frac{4}{13}$ then $\tan 2\alpha =?$

Answer 1

$\cos (\alpha +\beta )=3/5$,

$\sin (\alpha -\beta )=4/13$

$\alpha +\beta +\alpha -\beta =2\alpha$

$\tan (\alpha +\beta +\alpha -\beta )=\tan 2\alpha$

$\tan 2\alpha =\frac{\tan (\alpha +\beta )+\tan (\alpha -\beta )}{(1-\tan (\alpha +\beta \left)\tan \right(\alpha -\beta )}$

We have $\cos (\alpha +\beta )=\frac{3}{5}$

$\therefore \alpha +\beta$ is in I quadrant.

$\therefore \tan (\alpha +\beta )=\frac{4}{3}$

$\sin (\alpha -\beta )=\frac{4}{3}$

$\therefore \tan (\alpha -\beta )=\frac{4}{\sqrt{153}}$

$\therefore \tan 2\alpha =\frac{\frac{4}{3}+\frac{4}{\sqrt{152}}}{1-\frac{4}{13}\times \frac{4}{\sqrt{153}}}=\frac{4(\sqrt{153}+3)}{13\sqrt{153}-16}$.