Question

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

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

Answer 1

Answer: (A)

$cos \left(\alpha+\beta\right) = \frac{4}{5} \quad\Rightarrow tan \left(\alpha+\beta\right) = \frac{3}{4}$
$sin \left(\alpha-\beta\right) = \frac{5}{13}\quad\Rightarrow tan \left(\alpha-\beta\right) = \frac{5}{12}$
$tan \,2\alpha = tan\left(\alpha +\beta +\alpha -\beta \right) = \frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \frac{5}{12}} = \frac{56}{33}$