If $\cos (x-y),\cos x\text{ and }\cos (x+y)$ are in H.P., then $\left|\cos x\sec \frac{y}{2}\right|$ equal to 

It is given that 

$\cos (x-y),\cos x\text{ and }\cos (x+y)$ are in H.P

$\therefore \frac{2}{\cos x}=\frac{1}{\cos (x-y)}+\frac{1}{\cos (x+y)}$

$\begin{array}{l}\Rightarrow \frac{2}{\cos x}=\frac{2\cos x\cos y}{{\cos }^{2}x-{\sin }^{2}y}\\ \Rightarrow {\cos }^{2}x\cos y={\cos }^{2}x-{\sin }^{2}y\\ \Rightarrow {\cos }^{2}x(1-\cos y)={\sin }^{2}y\\ \Rightarrow 2{\cos }^{2}x{\sin }^2\frac{y}{2}=4{\sin }^2\frac{y}{2}{\cos }^2\frac{y}{2}\\ \Rightarrow {\cos }^{2}x{\sec }^2\frac{y}{2}=2\\ \Rightarrow \cos x\sec \frac{y}{2}\mid =\sqrt{2}\end{array}$