Given $0 < \alpha < {\pi \over 4}$ and $0 < \beta < {\pi \over 4}$ $\therefore$ $0 > - \beta > - {\pi \over 4}$ $\therefore$ $0 < \alpha + \beta < {\pi \over 2}$ and $- {\pi \over 4} < \alpha - \beta < {\pi \over 4}$ As cos($\alpha$ + $\beta$) = 3/5 so ${\tan \left( {\alpha + \beta } \right) = {4 \over 3}}$ As sin( $\alpha$ - $\beta$) = 5/13 so ${\tan \left( {\alpha - \beta } \right) = {5 \over {12}}}$ Now tan(2$\alpha$) = tan($\alpha$ + $\beta$ + $\alpha$ - $\beta$) = ${{\tan \left( {\alpha + \beta } \right) + \tan \left( {\alpha - \beta } \right)} \over {1 - \tan \left( {\alpha + \beta } \right)\tan \left( {\alpha - \beta } \right)}}$ = ${{{4 \over 3} + {5 \over {12}}} \over {1 - {4 \over 3} \times {5 \over {12}}}}$ = ${{63} \over {16}}$