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

  • question_answer
    If \[{{\sin }^{-1}}x+{{\cot }^{-1}}\left( \frac{1}{2} \right)=\frac{\pi }{2},\]then x is [Roorkee 1999; Karnataka CET 1999]

    A) 0

    B) \[\frac{1}{\sqrt{5}}\]

    C) \[\frac{2}{\sqrt{5}}\]

    D) \[\frac{\sqrt{3}}{2}\]

    Correct Answer: B

    Solution :

    \[\because \] \[{{\cot }^{-1}}\frac{1}{2}={{\cos }^{-1}}\frac{1}{\sqrt{5}}\] Hence given equation can be written as \[{{\sin }^{-1}}x+{{\cos }^{-1}}\frac{1}{\sqrt{5}}=\frac{\pi }{2}\] \[\Rightarrow \]\[x=\frac{1}{\sqrt{5}}\].


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