Here $\cos \alpha = {3 \over 5}$ $\therefore$ $\tan \alpha = {4 \over 3}$ and $\tan \beta = {1 \over 3}$ We know, $\tan \left( {\alpha - \beta } \right) = {{\tan \alpha - \tan \beta } \over {1 + \tan \alpha .\tan \beta }}$ = ${{{4 \over 3} - {1 \over 3}} \over {1 + {4 \over 3}.{1 \over 3}}}$ = ${9 \over {13}}$ $\therefore$ $\left( {\alpha - \beta } \right)$ = ${\tan ^{ - 1}}\left( {{9 \over {13}}} \right)$ = ${\sin ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)$