Question

If $\text{cos}\left(\alpha +\beta \right)=\frac{4}{5},\text{sin}\left(\alpha -\beta \right)=\frac{5}{13}\text{and}\alpha \text{,}\beta$ lie between $0$ and $\frac{\pi }{4},$ then $\text{tan 2}\alpha \text{=}$

Answer 1

We know that $2\alpha =(\alpha +\beta )+(\alpha -\beta )$

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

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

Given:$\alpha ,\beta$ lie between $0$ and $\frac{\pi }{4}$

$\therefore \alpha ,\beta$ lie in first quadrant.

$\Rightarrow \sin (\alpha +\beta )=\sqrt{1-{\cos }^2(\alpha +\beta )}=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{25-16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}$ in first quadrant.

$\Rightarrow \cos (\alpha -\beta )=\sqrt{1-{\sin }^2(\alpha -\beta )}=\sqrt{1-\frac{25}{169}}=\sqrt{\frac{169-25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13}$ in first quadrant.

$\Rightarrow \tan (\alpha -\beta )=\frac{\sin (\alpha -\beta )}{\cos (\alpha -\beta )}=\frac{\frac{5}{13}}{\frac{12}{13}}=\frac{5}{12}$

and $\tan (\alpha +\beta )=\frac{\sin (\alpha +\beta )}{\cos (\alpha +\beta )}=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$

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

$=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\times \frac{5}{12}}$

$=\frac{\frac{9+5}{12}}{\frac{48-15}{48}}$

$=\frac{\frac{14}{12}}{\frac{33}{48}}$

$=14\times \frac{4}{33}=\frac{56}{33}$

$\therefore \tan 2\alpha =\frac{56}{33}$