Question

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

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

Answer 1

Answer: (A)

$\frac{56}{33}$

$\tan 2\alpha =\frac{\sin 2\alpha }{\cos 2\alpha }$

$\Rightarrow \tan 2\alpha =\frac{\sin 2\alpha }{1-2{\sin }^{2}2\alpha }$

$\sin 2\alpha =sin\left[\right(\alpha +\beta )+(\alpha -\beta \left)\right]=\sin (\alpha +\beta )\cos (\alpha -\beta )+\sin (\alpha -\beta )\cos (\alpha +\beta )$

$=\frac{3}{5}\times \frac{12}{13}+\frac{5}{13}\times \frac{4}{5}=\frac{36}{65}+\frac{20}{65}=\frac{56}{65}$

$\cos 2\alpha =1-2{\sin }^{2}2\alpha =1-2{\left(\frac{56}{65}\right)}^2=\frac{33}{65}$

$\tan 2\alpha =\frac{\frac{56}{65}}{\frac{33}{65}}=\frac{56}{33}$