Question

If $\cos (\alpha +\beta )=\frac{4}{5},\sin (\alpha -\beta )=\frac{5}{13}$ and $\alpha ,\beta$ lie between 0 and $\frac{\pi }{4}$, then $\tan 2\alpha =$

• A. $\frac{16}{625}$
• B. $\frac{56}{33}$
• C. $\frac{3}{160}$
• D. $\frac{160}{3}$

Answer 1

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

It is given that $\alpha ,\beta$ lie between $0$ and $\frac{\pi }{4}$. Therefore, $-\pi /4<\alpha -\beta <\pi /4$ and $0<\alpha +\beta <\pi /2$.

So, $\cos (\alpha -\beta )$ and $\sin (\alpha +\beta )$ are positive.

Now, $\sin (\alpha +\beta )=\sqrt{1-{\cos }^2(\alpha +\beta )}⟹\sin (\alpha +\beta )=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$

and, $\cos (\alpha -\beta )=\sqrt{1-{\sin }^2(\alpha -\beta )}⟹\cos (\alpha -\beta )=\sqrt{1-\frac{25}{169}}=\frac{12}{13}$

Therefore,

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

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

Now, $\tan 2\alpha =\tan \left[\right(\alpha +\beta )+(\alpha -\beta \left)\right]=\frac{\tan (\alpha +\beta )+\tan (\alpha -\beta )}{1-\tan (\alpha +\beta )\tan (\alpha -\beta )}$

$=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\times \frac{5}{12}}=\frac{56}{33}$