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

  • question_answer
    If \[\cos (\theta -\alpha ),\ \cos \theta \]and \[\cos (\theta +\alpha )\]are in H.P., then \[\cos \theta \sec \frac{\alpha }{2}\]is equal to [IIT 1997]

    A) \[\pm \sqrt{2}\]

    B) \[\pm \sqrt{3}\]

    C) \[\pm 1/\sqrt{2}\]

    D) None of these

    Correct Answer: A

    Solution :

    Given \[\cos (\theta -\alpha ),\cos \theta \]and \[\cos (\theta +\alpha )\]are in H.P. Þ \[\frac{1}{\cos (\theta -\alpha )},\frac{1}{\cos \theta },\frac{1}{\cos (\theta +\alpha )}\]will be in A.P. Hence, \[\frac{2}{\cos \theta }=\frac{1}{\cos (\theta -\alpha )}+\frac{1}{\cos (\theta +\alpha )}\]     \[=\frac{\cos (\alpha +\theta )+\cos (\theta -\alpha )}{{{\cos }^{2}}\theta -{{\sin }^{2}}\alpha }\] Þ \[\frac{2}{\cos \theta }=\frac{2\cos \theta \cos \alpha }{{{\cos }^{2}}\theta -{{\sin }^{2}}\alpha }\] Þ \[{{\cos }^{2}}\theta -{{\sin }^{2}}\alpha ={{\cos }^{2}}\theta \cos \alpha \] Þ \[{{\cos }^{2}}\theta \,(1-\cos \alpha )={{\sin }^{2}}\alpha \] Þ \[{{\cos }^{2}}\theta \left( 2{{\sin }^{2}}\frac{\alpha }{2} \right)=4{{\sin }^{2}}\frac{\alpha }{2}{{\cos }^{2}}\frac{\alpha }{2}\] \[{{\cos }^{2}}\theta {{\sec }^{2}}\frac{\alpha }{2}=2\Rightarrow \cos \theta \sec \frac{\alpha }{2}=\pm \sqrt{2}\].

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