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 $\tan 2\alpha =$

Answer 1

Given: $\cos (\alpha +\beta )=\frac{4}{5}\text{and}\sin (\alpha –\beta )=\frac{5}{13}$

Using ${\sin }^2\theta +{\cos }^2\theta =1$, we get
$\sin (\alpha +\beta )=\frac{3}{5}\text{and}\cos (\alpha –\beta )=\frac{12}{13}$
So we can say, $\tan (\alpha +\beta )=\frac{\sin (\alpha +\beta )}{cos(\alpha +\beta )}=\frac{\left(3/5\right)}{\left(4/5\right)}=\frac{3}{4}$
and $\tan (\alpha –\beta )=\frac{\sin (\alpha –\beta )}{\cos (\alpha –\beta )}=\frac{\left(5/13\right)}{\left(12/13\right)}=\frac{5}{12}$

Taking $\tan 2\alpha =\tan (\alpha +\beta +\alpha –\beta )$
$\Rightarrow \tan 2\alpha =\frac{\tan (\alpha +\beta )+\tan (\alpha –\beta )}{1–\tan (\alpha +\beta )\tan (\alpha –\beta )}$

Now substituting thes values from above,
$\Rightarrow \tan 2\alpha =\frac{\left(\frac{3}{4}\right)+\left(\frac{5}{12}\right)}{1–\left(\frac{3}{4}\right)\left(\frac{5}{12}\right)}=\frac{\frac{9+5}{12}}{\frac{48–15}{48}}=\frac{56}{33}$