Question

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

Answer 1

Given:
$\cos (\alpha +\beta )=\frac{4}{5},\sin (\alpha -\beta )=\frac{5}{13}\dots \left(1\right)$
$\sin (\alpha +\beta )=\sqrt{1-{\cos }^2(\alpha +\beta })$
$[\because {\sin }^{2}x+{\cos }^{2}x=1]$
$=\sqrt{1-\frac{16}{25}}$
$=\sqrt{\frac{9}{25}}$
$\sin (\alpha +\beta )=\frac{3}{5}\dots \left(2\right)$
$and\cos (\alpha -\beta )=\sqrt{1-si{n}^2(\alpha -\beta )}$
$[\because {\sin }^{2}x+co{s}^{2}x=1]$
$=\sqrt{1-\frac{25}{169}}$
$=\sqrt{\frac{144}{169}}$
$\cos (\alpha -\beta )=\frac{12}{13}\dots \left(3\right)$
From equations $\left(1\right),\left(2\right)$ and $\left(3\right)$
$\tan (\alpha +\beta )=\frac{\sin (\alpha +\beta )}{\cos (\alpha +\beta )}=\frac{\frac{3}{5}}{\frac{4}{5}}$
$\Rightarrow \tan (\alpha +\beta )=\frac{3}{4}\dots \left(4\right)$
$and\tan (\alpha -\beta )=\frac{\sin (\alpha -\beta )}{\cos (\alpha -\beta )}=\frac{\frac{5}{13}}{\frac{12}{13}}$
$\Rightarrow \tan (\alpha -\beta )=\frac{5}{12}\dots \left(5\right)$
$\tan 2\alpha =\tan \left[\right(\alpha +\beta )+(\alpha -\beta \left)\right]$
$=\frac{\tan (\alpha +\beta )+\tan (\alpha -\beta )}{1-\tan (\alpha +\beta )\tan (\alpha -\beta )}$
$=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\times \frac{5}{12}}(\text{From}eq.(4\left)\text{and}\right(5\left)\right)$
$=\frac{\frac{14}{12}}{1-\frac{15}{48}}=\frac{\frac{14}{12}}{\frac{33}{48}}=\frac{14}{12}\times \frac{48}{33}$
$\tan 2\alpha =\frac{56}{33}$