JEE Main 2021Inverse Trigonometric FunctionsInverse Trigonometric FunctionsEasyQuestionThe value of sin−1(1213)−sin−1(35){\sin ^{ - 1}}\left( {{{12} \over {13}}} \right) - {\sin ^{ - 1}}\left( {{3 \over 5}} \right)sin−1(1312)−sin−1(53) is equal to :OptionsAπ−sin−1(6365)\pi - {\sin ^{ - 1}}\left( {{{63} \over {65}}} \right)π−sin−1(6563)Bπ2−sin−1(5665){\pi \over 2} - {\sin ^{ - 1}}\left( {{{56} \over {65}}} \right)2π−sin−1(6556)Cπ2−cos−1(965){\pi \over 2} - {\cos ^{ - 1}}\left( {{9 \over {65}}} \right)2π−cos−1(659)Dπ−cos−1(3365)\pi - {\cos ^{ - 1}}\left( {{{33} \over {65}}} \right)π−cos−1(6533)Check AnswerHide SolutionSolutionsin−11213−sin−135=sin−1(1213.45.35.513){\sin ^{ - 1}}{{12} \over {13}} - {\sin ^{ - 1}}{3 \over 5} = {\sin ^{ - 1}}\left( {{{12} \over {13}}.{4 \over 5}.{3 \over 5}.{5 \over {13}}} \right)sin−11312−sin−153=sin−1(1312.54.53.135) ⇒sin−13365=π2−cos−13365 \Rightarrow {\sin ^{ - 1}}{{33} \over {65}} = {\pi \over 2} - {\cos ^{ - 1}}{{33} \over {65}}⇒sin−16533=2π−cos−16533 ⇒π2−sin−15665 \Rightarrow {\pi \over 2} - {\sin ^{ - 1}}{{56} \over {65}}⇒2π−sin−16556