If $\cos (\theta -\alpha ),\cos \theta ,\cos (\theta +\alpha )$ are in H.P., then $\cos \theta \sec (\alpha /2)$is equal to 

We are given that 

$\begin{array}{r}\cos \theta =\frac{2\cos (\theta -\alpha )\cos (\theta +\alpha )}{\cos (\theta -\alpha )+\cos (\theta +\alpha )}\\ \Rightarrow \cos \theta =\frac{2\left({\cos }^2\theta -{\sin }^2\alpha \right)}{2\cos \theta \cos \alpha }\end{array}$

$\begin{array}{l}\Rightarrow {\cos }^2\theta \cos \alpha ={\cos }^2\theta -{\sin }^2\alpha \\ \Rightarrow {\cos }^2\theta =\frac{{\sin }^2\alpha }{1-\cos \alpha }\\ \Rightarrow \cos \theta =\frac{4{\sin }^2\frac{\alpha }{2}{\cos }^2\frac{\alpha }{2}}{2{\sin }^2\frac{\alpha }{2}}\\ \Rightarrow {\cos }^2\theta {\sec }^2\frac{\alpha }{2}=2\Rightarrow \cos \theta \sec \frac{\alpha }{2}=\pm \sqrt{2}\end{array}$