If $0<A<B<\frac{\pi }{4}$  and $\sin (A+B)=\frac{24}{25}$   and  $\cos (A-B)=\frac{4}{5}$  then find the value of tan $2A.$

$O<A<B<\frac{\pi }{4}\Rightarrow 0<(A+B)<\frac{\pi }{2}\Rightarrow A+B\in Q_1$

Since $A<B\Rightarrow A-B<0\Rightarrow A-B\in Q_4$

$\mathrm{Sin}(A+B)=\frac{24}{25} \mathrm{Cos}(A-B)=\frac{4}{5}$

$\mathrm{Tan}(A+B)=\frac{24}{7}$, $\mathrm{Tan}(A-B)=-\frac{3}{4}$

$\mathrm{Tan}2A=\mathrm{Tan}\left[\right(A+B)+(A-B\left)\right]$

$=\frac{\mathrm{Tan}(A+B)+\mathrm{Ta}n(A-B)}{1-\mathrm{Tan}(A+B)\mathrm{Tan}(A-B)}$

$=\frac{\frac{24}{7}+\left(\frac{-3}{4}\right)}{1-\frac{24}{7}\left(\frac{-3}{4}\right)}=\frac{\frac{24}{7}-\frac{3}{4}}{1+\frac{72}{28}}$

$=\frac{\frac{96-21}{28}}{\underset{ }{28+72}}=\frac{75}{100}=\frac{3}{4}$

$\therefore$$\mathrm{Tan}2A=\frac{3}{4}$