Trigonometry! 12

Using Pythagoras Theorem:-
$Sin(\alpha +\beta)=\dfrac{3}{5}, ~and ~~~~~~~~Cos(\alpha -\beta)=\dfrac{12}{13} \\Expanding~~\\Sin\alpha*Cos\beta+Cos\alpha*Sin\beta =\dfrac{3}{5} ,~~~Sin\alpha*Cos\beta-Cos\alpha*Sin\beta =\dfrac{5}{13} \\ Cos\alpha*Cos\beta-Sin\alpha*Sin\beta =\dfrac{4}{5} ,~~~Cos\alpha*Cos\beta+Sin\alpha*Sin\beta =\dfrac{12}{13}\\ \implies~2Sin\alpha*Cos\beta=\dfrac{3}{5}+ \dfrac{5}{13}=\dfrac{64}{65} ...(1) , ~and~~~~~\\\implies~2Cos\alpha*Cos\beta=\dfrac{4}{5}+\dfrac{12}{13}=\dfrac{112}{65}.......(2)\\\therefore~ \dfrac{(1)}{(2)}=tan(\alpha)=\dfrac{64}{112}=\dfrac{4}{7}\\\therefore~tan2\alpha=\dfrac{2tan(\alpha)}{1-tan^2\alpha }=~~~~\color{#D61F06}{\boxed{\dfrac{56}{33} } }$

$\begin{aligned} \tan 2 \alpha & = \frac {\sin 2 \alpha}{\cos 2 \alpha} \\ & = \frac {\sin ((\alpha + \beta) +(\alpha - \beta))}{\cos ((\alpha + \beta) +(\alpha - \beta))} \\ & = \frac {\sin(\alpha + \beta)\cos(\alpha - \beta) + \cos(\alpha + \beta)\sin(\alpha - \beta)}{\cos(\alpha + \beta)\cos(\alpha - \beta) - \sin(\alpha + \beta)\sin(\alpha - \beta)} \\ & = \frac {\frac 35 \times \frac {12}{13} + \frac 45 \times \frac {5}{13}}{\frac 45 \times \frac {12}{13} - \frac 35 \times \frac {5}{13}} = \frac {36+20}{48-15} = \boxed{\dfrac {56}{33}} \end{aligned}$

Well, I tried a long method. I dont have time. sry, sin(2α) = sin[(α+ß)-(α-ß)] , cos(α-ß) = 12/13, sin(α-ß) = 3/5 Similarly, do that for cos(2α) And youll get after proper simplifications. Very easy. Please ask if explanation needed, but miene is a very long one :D