JEE Main & Advanced Mathematics Trigonometric Identities Question Bank Trigonometrical ratios of sum and difference of two and three angles

  • question_answer
    \[\frac{\sin 3\theta +\sin 5\theta +\sin 7\theta +\sin 9\theta }{\cos 3\theta +\cos 5\theta +\cos 7\theta +\cos 9\theta }=\] [Roorkee 1973]

    A) \[\tan 3\theta \]

    B) \[\cot 3\theta \]

    C) \[\tan 6\theta \]

    D) \[\cot 6\theta \]

    Correct Answer: C

    Solution :

    \[\frac{\sin \,\,3\theta +\sin \,\,5\theta +\sin \,7\theta +\sin 9\theta }{\cos 3\theta +\cos 5\theta +\cos \,7\theta +\cos \,9\theta }\] \[=\frac{(\sin \,3\theta +\sin \,9\theta )+(\sin \,5\theta +\sin \,7\theta )}{(\cos \,3\theta +\cos \,9\theta )+(\cos \,5\theta +\cos \,7\theta )}\] \[=\frac{2\,\sin \,6\theta \,\cos \,3\theta +2\,\sin \,6\theta \,\cos \,\theta }{2\,\cos \,6\theta \,\cos \,3\theta +2\,\cos \,6\theta \,\cos \,\theta }\] \[=\frac{2\,\sin \,6\theta \,(\cos \,3\theta +\cos \theta )}{2\,\cos \,6\theta \,(\cos \,3\theta +\cos \theta )}=\tan \,6\theta \].


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