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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