If \alpha = {\cos ^{ - 1}}\left( {{3 \over 5}} ight), \beta = { an ^{ - 1}}\left( {{1 \over 3}} ight) where 0

Question

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

Accepted Answer

Answer: B. {\sin ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} ight) Here \cos \alpha = {3 \over 5} 	herefore 	an \alpha = {4 \over 3} and 	an \beta = {1 \over 3} We know, 	an \left( {\alpha - \beta } ight) = {{	an \alpha - 	an \beta } \over {1 + 	an \alpha .	an \beta }} = {{{4 \over 3} - {1 \over 3}} \over {1 + {4 \over 3}.{1 \over 3}}} = {9 \over {13}} 	herefore \left( {\alpha - \beta } ight) = {	an ^{ - 1}}\left( {{9 \over {13}}} ight) = {\sin ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} ight)

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 :

Answer

Explanation

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