JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Morning)

  • question_answer
    The value of \[{{\sin }^{-1}}\left( \frac{12}{13} \right)-{{\sin }^{-1}}\left( \frac{3}{5} \right)\]is equal to:                        [JEE Main Held on 12-4-2019 Morning]

    A) \[\pi -{{\sin }^{-1}}\left( \frac{63}{65} \right)\]      

    B) \[\pi -{{\cos }^{-1}}\left( \frac{33}{65} \right)\]

    C) \[\frac{\pi }{2}-{{\sin }^{-1}}\left( \frac{56}{65} \right)\]

    D) \[\frac{\pi }{2}-{{\cos }^{-1}}\left( \frac{9}{65} \right)\]

    Correct Answer: C

    Solution :

    \[{{\sin }^{-1}}\left( \frac{12}{13} \right)-{{\sin }^{-1}}\left( \frac{3}{5} \right)\]           \[{{\sin }^{-1}}\left( x\sqrt{1-{{y}^{2}}}-y\sqrt{1-{{x}^{2}}} \right)\]           \[={{\sin }^{-1}}\left( \frac{33}{65} \right)={{\cos }^{-1}}\left( \frac{56}{65} \right)=\frac{\pi }{2}-{{\sin }^{1}}\left( \frac{56}{65} \right)\]      


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