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JEE Main & Advanced Mathematics Inverse Trigonometric Functions Question Bank Self Evaluation Test - Inverse Trigonometric Functions

  • question_answer
    \[\tan \left\{ \frac{1}{2}{{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}+\frac{1}{2}{{\cos }^{-1}}\frac{1-{{y}^{2}}}{1+{{y}^{2}}} \right\}=\]

    A) \[\frac{x-y}{1+xy}\]

    B) \[\frac{x+y}{1-xy}\]

    C) \[\frac{x-y}{x+y}\]

    D) \[\frac{1-xy}{1+xy}\]

    Correct Answer: B

    Solution :

    [b] Put \[x=\tan \theta \] and \[y=\tan \phi \]


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