• # 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

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 ${{\cos }^{2}}\theta \left( 2{{\sin }^{2}}\frac{\alpha }{2} \right)=4{{\sin }^{2}}\frac{\alpha }{2}{{\cos }^{2}}\frac{\alpha }{2}$ ${{\cos }^{2}}\theta {{\sec }^{2}}\frac{\alpha }{2}=2\Rightarrow \cos \theta \sec \frac{\alpha }{2}=\pm \sqrt{2}$.