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

  • question_answer
    The solution set of the equation \[{{\sin }^{-1}}x=2{{\tan }^{-1}}x\] is [AMU 2002]

    A) {1, 2}

    B) {-1, 2}

    C) {-1, 1, 0}

    D) {1, 1/2, 0}

    Correct Answer: C

    Solution :

      \[{{\sin }^{-1}}x=2{{\tan }^{-1}}x\] Þ \[{{\sin }^{-1}}x={{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}\] \[\Rightarrow \frac{2x}{1+{{x}^{2}}}=x\]              Þ \[{{x}^{3}}-x=0\] Þ  \[x(x+1)(x-1)=0\]  Þ \[x=\left\{ -1,\,\,1,\,\,0 \right\}\].


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