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)

$cos\left(\alpha +\beta \right)=\frac{4}{5}\Rightarrow tan\left(\alpha +\beta \right)=\frac{3}{4}$
$sin\left(\alpha -\beta \right)=\frac{5}{13}\Rightarrow tan\left(\alpha -\beta \right)=\frac{5}{12}$
$tan2\alpha =tan\left(\alpha +\beta +\alpha -\beta \right)=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\frac{5}{12}}=\frac{56}{33}$