Question

Let $\cos (\alpha +\beta )=\frac{4}{5}$ and let $\sin (\alpha -\beta )=\frac{5}{13}$ where $0\le \alpha ,\beta \le \frac{\pi }{4}$, then $\tan 2\alpha =$

• A. $\frac{25}{16}$
• B. $\frac{56}{33}$
• C. $\frac{19}{12}$
• D. $\frac{20}{7}$

Answer 1

The correct option is B $\frac{56}{33}$

$\cos (\alpha +\beta )=\frac{4}{5}$ $\sin (\alpha -\beta )=\frac{5}{13}$

$\sin (\alpha +\beta )=\frac{3}{5}$ $\cos (\alpha -\beta )=\frac{12}{13}$

$\sin (\alpha +\beta )+\sin (\alpha -\beta )=2\sin \alpha \cos \beta$

$\frac{5}{13}+\frac{3}{5}=\frac{64}{65}$

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

$=\frac{4}{5}+\frac{2}{13}=\frac{112}{65}$

$\frac{2\sin \alpha \cos \beta }{2\cos \alpha \cos \beta }=\tan \alpha =\frac{\frac{64}{65}}{\frac{112}{65}}=\frac{4}{7}$

$\tan \left(2\alpha \right)=\frac{2\tan \left(\alpha \right)}{1-{\tan }^2\left(\alpha \right)}=\frac{2\times \frac{4}{7}}{1-\frac{16}{49}}$

$\tan \left(2\alpha \right)=\frac{\frac{8}{7}}{\frac{33}{49}}=\frac{56}{33}$.