JEE Main & Advanced Mathematics Inverse Trigonometric Functions Question Bank Inverse trigonometric functions

  • question_answer
    \[\cos \left[ 2{{\cos }^{-1}}\frac{1}{5}+{{\sin }^{-1}}\frac{1}{5} \right]=\] [IIT 1981]

    A) \[\frac{2\sqrt{6}}{5}\]

    B) \[-\frac{2\sqrt{6}}{5}\]

    C) \[\frac{1}{5}\]

    D) \[-\frac{1}{5}\]

    Correct Answer: B

    Solution :

      \[\cos \,\left( {{\cos }^{-1}}\frac{1}{5}+{{\sin }^{-1}}\frac{1}{5}+{{\cos }^{-1}}\frac{1}{5} \right)=\cos \,\left( \frac{\pi }{2}+{{\cos }^{-1}}\frac{1}{5} \right)\] \[=-\sin \,\left( {{\cos }^{-1}}\frac{1}{5} \right)=-\sin \,\left( {{\sin }^{-1}}\sqrt{\frac{24}{25}} \right)=-\frac{2\sqrt{6}}{5}\].


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