Question

If $0,\alpha ,\beta <\frac{\pi }{4}$ such that $\cos (\alpha +\beta )=\frac{4}{5}$ and$\sin (\alpha -\beta )=\frac{5}{13}$, then the value of $\tan 2\alpha =$

• A. $\frac{56}{65}$
• B. $\frac{56}{23}$
• C. $\frac{56}{33}$
• D. $\frac{56}{36}$

Answer 1

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

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

$\sin (\alpha +\beta )=\sqrt{1-{\left(\frac{4}{5}\right)}^2}$ $0<\alpha ,\beta <\pi /4$

$=\sqrt{1-\frac{16}{25}}=3/5$

$\sin (\alpha -\beta )=5/13$

$\therefore \cos (\alpha -\beta )=12/13$

$\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta =3/5$

$\sin (\alpha -\beta )=\frac{\sin \alpha \cos \beta -\cos \alpha \sin \beta =5/13}{2\sin \alpha \cos \beta =\frac{3}{5}+\frac{5}{13}=\frac{64}{65}}$

$\sin \alpha \cos \beta =32/65$

$\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta =4/5$

$\cos (\alpha -\beta )=\frac{\cos \alpha \cos \beta +\sin \alpha \sin \beta =12/13}{2\cos \alpha \cos \beta =\frac{4}{5}+\frac{12}{13}}=\frac{52+60}{65}=\frac{112}{65}$

$\cos \alpha \cos \beta =\frac{56}{65}$...........$\left(2\right)$

$\left(1\right)÷\left(2\right)\Rightarrow \tan \alpha =32/56=4/7$

$\tan 2\alpha =\frac{2\tan \alpha }{1-{\tan }^2\alpha }=\frac{2.4/7}{1-16/49}=\frac{8/7}{33/49}$

$=\frac{8\times 497}{7\times 33}=\frac{56}{33}$