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

  • question_answer
    If\[\frac{2\sin \alpha }{\{1+\cos \alpha +\sin \alpha \}}=y,\]then \[\frac{\{1-\cos \alpha +\sin \alpha \}}{1+\sin \alpha }=\] [BIT Ranchi 1996; Orissa JEE 2004]

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

    B) \[y\]

    C) \[1-y\]

    D) \[1+y\]

    Correct Answer: B

    Solution :

    We have, \[\frac{2\sin \alpha }{1+\cos \alpha +\sin \alpha }=y\] Then \[\frac{4\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}{2{{\cos }^{2}}\frac{\alpha }{2}+2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}=y\] Þ \[\frac{2\sin \frac{\alpha }{2}}{\cos \frac{\alpha }{2}+\sin \frac{\alpha }{2}}\times \frac{\left( \sin \frac{\alpha }{2}+\cos \frac{\alpha }{2} \right)}{\left( \sin \frac{\alpha }{2}+\cos \frac{\alpha }{2} \right)}=y\] Þ \[\frac{1-\cos \alpha +\sin \alpha }{1+\sin \alpha }=y\].

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