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

Answer 1

Answer: (A)

$\frac{56}{33}$

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

Now,

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

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

$\Rightarrow \sin 2\alpha =\frac{3}{5}.\frac{12}{13}+\frac{4}{5}.\frac{5}{13}$

$=\frac{36}{65}+\frac{20}{65}=\frac{56}{65}$

$\cos 2\alpha =\frac{4}{5}\times \frac{12}{13}-\frac{3}{5}\times \frac{5}{13}$

$=\frac{48-15}{65}=\frac{33}{65}$

$\therefore$ $\tan 2\alpha =\frac{\sin 2\alpha }{\cos 2\alpha }=\frac{56}{33}$ [A]