Question

lf $\cos (\theta -\alpha ),\cos \theta ,\cos (\theta +\alpha )$ are in $H.P$. then $\cos \theta .\sec \frac{\alpha }{2}=$

• A. $-\sqrt{2}$
• B. $\sqrt{2}$
• C. $\pm \sqrt{2}$
• D. $1$

Answer 1

Answer: (C)

$\pm \sqrt{2}$

$\cos (\theta -\alpha ),\cos \theta ,\cos (\theta +\alpha )$ are in H.P.
$\Rightarrow \frac{2\cos (\theta -\alpha ).\cos (\theta +\alpha )}{\cos (\theta -\alpha )+\cos (\theta +\alpha )}=\cos \theta$
$\Rightarrow \frac{2({\cos }^2\theta -{\sin }^2\alpha )}{2\cos \theta .\cos \alpha }=\cos \theta$
$\Rightarrow 2{\cos }^2\theta -2{\sin }^2\alpha =2{\cos }^2\theta \cos \alpha \Rightarrow 2{\cos }^2\theta (1-\cos \alpha )=2{\sin }^2\alpha \Rightarrow 2{\cos }^2\theta .2{\sin }^2\left(\alpha /2\right)=8{\sin }^2\left(\alpha /2\right){\cos }^2\left(\alpha /2\right)\Rightarrow {\cos }^2\theta .{\sec }^2\left(\alpha /2\right)=2\Rightarrow \cos \theta .\sec \left(\alpha /2\right)=\pm \sqrt{2}$