Explanation

(b)
We have, $\cos(\alpha + \beta) = \frac{4}{5}\ and\sin{(\alpha - \beta) = \frac{5}{13}}$
$\Rightarrow \sin{(\alpha + \beta) = \frac{3}{5}\ and\cos{(\alpha - \beta) = \frac{12}{13}}}$
$\Rightarrow (\alpha + \beta) = \sin^{- 1}{\frac{3}{5}\ and\ (\alpha - \beta) = \sin^{- 1}\frac{5}{13}}$
$\therefore 2\alpha = \sin^{- 1}{\frac{3}{5} + \sin^{- 1}\frac{5}{13}}$
$= \sin^{- 1}\left\lbrack \frac{3}{5}\sqrt{1 - \frac{25}{169} + \frac{5}{13}}\sqrt{1 - \frac{9}{25}} \right\rbrack$
$= \sin^{- 1}{\left( \frac{3}{5} \times \frac{12}{13} + \frac{5}{13} \times \frac{4}{5} \right) = \sin^{- 1}\left( \frac{36}{65} + \frac{20}{65} \right)}$
$\Rightarrow \sin{2\alpha = \frac{56}{65}}$
$\therefore\tan{2\alpha = \frac{56}{33}}$