JEE Main & Advanced Mathematics Trigonometric Identities Question Bank Trigonometrical ratios of multiple and sub multiple angles

  • question_answer
    \[\tan \frac{A}{2}\]is equal to

    A) \[\pm \sqrt{\frac{1-\sin A}{1+\sin A}}\]

    B) \[\pm \sqrt{\frac{1+\sin A}{1-\sin A}}\]

    C) \[\pm \sqrt{\frac{1-\cos A}{1+\cos A}}\]

    D) \[\pm \sqrt{\frac{1+\cos A}{1-\cos A}}\]

    Correct Answer: C

    Solution :

    \[\tan \left( \frac{A}{2} \right)=\frac{\sin (A/2)}{\cos (A/2)}=\pm \sqrt{\frac{(1-\cos A)/2}{(1+\cos A)/2}}=\pm \sqrt{\frac{1-\cos A}{1+\cos A}}\].

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