If $\alpha = {\cos ^{ - 1}}\left( {{3 \over 5}} \right)$, $\beta = {\tan ^{ - 1}}\left( {{1 \over 3}} \right)$ where $0 < \alpha ,\beta < {\pi \over 2}$ , then $\alpha $ - $\beta $ is equal to :

Question

${\tan ^{ - 1}}\left( {{9 \over {14 }}} \right)$

${\sin ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)$

${\cos ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)$

${\tan ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)$

Solution

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