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

  • question_answer
    . Given that \[\cos \left( \frac{\alpha -\beta }{2} \right)=2\cos \left( \frac{\alpha +B}{2} \right)\], then \[\tan \frac{\alpha }{2}\tan \frac{\beta }{2}\]is equal to [AMU 2001]

    A) \[\frac{1}{2}\]

    B) \[\frac{1}{3}\]

    C) \[\frac{1}{4}\]

    D) \[\frac{1}{8}\]

    Correct Answer: B

    Solution :

    \[\cos \left( \frac{\alpha -\beta }{2} \right)=2\cos \left( \frac{\alpha +\beta }{2} \right)\] Þ \[\cos \frac{\alpha }{2}\cos \frac{\beta }{2}+\sin \frac{\alpha }{2}\sin \frac{\beta }{2}=2\cos \frac{\alpha }{2}\cos \frac{\beta }{2}-2\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\]         \[\Rightarrow 3\sin \frac{\alpha }{2}\sin \frac{\beta }{2}=\cos \frac{\alpha }{2}\cos \frac{\beta }{2}\] Þ \[\tan \frac{\alpha }{2}\tan \frac{\beta }{2}=\frac{1}{3}\].

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