Question

If $\cos (\alpha +\beta )=\frac{3}{5},\sin (\alpha -\beta )=\frac{5}{13}$ and $0<\alpha ,\beta <\frac{\pi }{4}$, then $\tan \left(2\alpha \right)$ is equal to:

• A. $\frac{21}{16}$
• B. $\frac{63}{52}$
• C. $\frac{33}{52}$
• D. $\frac{63}{16}$

Answer 1

Answer: (D)

$\frac{63}{16}$

$0<\alpha +\beta =\frac{\pi }{2}$ and $\frac{-\pi }{4}<\alpha -\beta <\frac{\pi }{4}$
if $\cos (\alpha +\beta )=\frac{3}{5}$ then $\tan (\alpha +\beta )=\frac{4}{3}$
and if $\sin (\alpha -\beta )=\frac{5}{12}$
(since $\alpha -\beta$ here lies in the first quadrant)
Now $\tan \left(2\alpha \right)=\tan \left\{\right(a+\beta )+(\alpha -\beta \left)\right\}$
$=\frac{\tan (\alpha +\beta )+\tan (\alpha -\beta )}{1-\tan (\alpha +\beta ).\tan (\alpha -\beta )}=\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3}.\frac{5}{12}}=\frac{63}{16}$