{\sin ^1}\left( {\sin {{2\pi } \over 3}} ight) + {\cos ^{ - 1}}\left( {\cos {{7\pi } \over 6}} ight) + { an ^{ - 1}}\left( { an {{3\pi } \over 4}} ight) is equal to :

Question

{\sin ^1}\left( {\sin {{2\pi } \over 3}} ight) + {\cos ^{ - 1}}\left( {\cos {{7\pi } \over 6}} ight) + {	an ^{ - 1}}\left( {	an {{3\pi } \over 4}} ight) is equal to :

Accepted Answer

Answer: A. {{11\pi } \over {12}} {\sin ^{ - 1}}\left( {{{\sqrt 3 } \over 2}} ight) + {\cos ^{ - 1}}\left( {{{ - \sqrt 3 } \over 2}} ight) + {	an ^{ - 1}}\left( { - 1} ight) = {\pi \over 3} + {{5\pi } \over 6} - {\pi \over 4} = {{4\pi + 10\pi - 3\pi } \over {12}} = {{11\pi } \over {12}}

$${\sin ^1}\left( {\sin {{2\pi } \over 3}} \right) + {\cos ^{ - 1}}\left( {\cos {{7\pi } \over 6}} \right) + {\tan ^{ - 1}}\left( {\tan {{3\pi } \over 4}} \right)$$ is equal to :

Answer

Explanation

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