If $\cos (x-y),\cos x$ and $\cos (x+y)$ are in $\mathrm{HP}$, then $\cos x\sec (y/2)$ is equal to

Given that, $\cos (x-y),\cos x,\cos (x+y)$ are in $\mathrm{HP}$.

Then, $\cos x=\frac{2\cos (x-y)\cos (x+y)}{\cos (x-y)+\cos (x+y)}$

$\Rightarrow \cos x=\frac{2\left({\cos }^{2}x-{\sin }^{2}y\right)}{2\cos x\cos y}$

$\Rightarrow {\cos }^{2}x\cos y={\cos }^{2}x-{\sin }^{2}y$

$\Rightarrow {\cos }^{2}x(\cos y-1)=-{\sin }^{2}y$

$\Rightarrow {\cos }^{2}x(1-\cos y)=1-{\cos }^{2}y$

$\Rightarrow {\cos }^{2}x(1-\cos y)=(1-\cos y)(1+\cos y)$

$\Rightarrow {\cos }^{2}x=1+\cos y\Rightarrow {\cos }^{2}x=2{\cos }^{2}y/2$

$\Rightarrow {\cos }^{2}x{\sec }^2(y/2)=2$

$\Rightarrow \cos x\sec (y/2)=\pm \sqrt{2}$