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

  • question_answer
    \[{{\sin }^{-1}}\left[ x\sqrt{1-x}-\sqrt{x}\sqrt{1-{{x}^{2}}} \right]=\]

    A) \[{{\sin }^{-1}}x+{{\sin }^{-1}}\sqrt{x}\]

    B) \[{{\sin }^{-1}}x-{{\sin }^{-1}}\sqrt{x}\]

    C) \[{{\sin }^{-1}}\sqrt{x}-{{\sin }^{-1}}x\]

    D) None of these

    Correct Answer: B

    Solution :

      Let \[x=\sin \theta \] and \[\sqrt{x}=\sin \varphi \] Hence \[{{\sin }^{-1}}(x\sqrt{1-x}-\sqrt{x}\,\sqrt{1-{{x}^{2}}})\] \[={{\sin }^{-1}}(\sin \theta \sqrt{1-{{\sin }^{2}}\varphi }-\sin \varphi \sqrt{1-{{\sin }^{2}}\theta })\] \[={{\sin }^{-1}}(\sin \theta \cos \varphi -\sin \varphi \cos \theta )={{\sin }^{-1}}\sin \,(\theta -\varphi )\] \[=\theta -\varphi ={{\sin }^{-1}}(x)-{{\sin }^{-1}}(\sqrt{x})\].


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