If sec \((\mathrm{x}-\mathrm{y}), \sec \mathrm{x}\) and \(\mathrm{sec}(\mathrm{x}+\mathrm{y})\) are in A. P., then \(\cos x \sec (y / 2)=\ldots \ldots \ldots(y \neq 2 n \pi, n \in I)\) \(\begin{array}{llll}\text { (A) } \pm \sqrt{2} & \text { (B) } \pm(1 / \sqrt{2}) & \text { (C) } \pm 2 & \text { (D) } \pm(1 / 2)\end{array}\)

Since sec(x-y), sec(x), and sec(x+y) are in A.P, the middle term sec(x) is the average of the other two terms: \[sec(x)=\frac{sec(x-y)+sec(x+y)}{2}\]

We will use the trigonometric identity \(secA+secA=2secA(1+\tan{A}\tan{(-A)})\) to simplify the expression. \[2sec(x)=sec(x-y)+sec(x+y)=sec(x-y)+sec(x+y)-2sec(x)+2sec(x)\] \[2sec(x)=2sec(x)\left[1+\tan{(x-y)}\tan{(x+y)}\right]\] Now, divide both sides by \(2sec(x)\): \[1=\tan{(x-y)}\tan{(x+y)}\] Next, apply the tangent addition formula on the RHS: \[\tan{(x-y)}\tan{(x+y)}=\frac{\sin{(x-y)}\sin{(x+y)}}{\cos{(x-y)}\cos{(x+y)}}\] Applying the sine and the cosine angle addition formulas, we get: \[\frac{(\sin x \cos y - \cos x \sin y)(\sin x \cos y + \cos x \sin y)}{(\cos x \cos y + \sin x \sin y)(\cos x \cos y - \sin x \sin y)}=\frac{(\sin^2{x}\cos^2{y}-\cos^2{x}\sin^2{y})}{1-\sin^{2}{x}\sin^{2}{y}}=1\] Now, subtract 1 from both sides, and utilize the Pythagorean identity \(\sin^2{u}+\cos^2{u}=1\): \[\frac{\sin^2{x}\cos^2{y}+\cos^2{x}\sin^2{y}-2\sin^2{x} \cos^2{y}}{1-\sin^{2}{x}\sin^{2}{y}}=0\] This simplifies to: \[\cos^2{x}\cos^2{y}-\sin^2{x}\sin^2{y}=0\] \[\cos^2{x}\cos^2{y}=\sin^2{x}\sin^2{y}\]

Divide both sides by \(\cos^2{x}\cos^2{y}\) and use the identity \(\sec^2{u}=1+\tan^2{u}\): \[\tan^2{x}\tan^2{y}=1\] Now, notice that \(\tan^2{y/2}=\frac{1-\cos{y}}{1+\cos{y}}\). Therefore, \[\tan^2{x}\frac{1-\cos{y}}{1+\cos{y}}=1\] \[\tan^2{x}(1-\cos y)=1+\cos y\] Multiply both sides by \(\cos^2{x}\) and apply the Pythagorean identity \(\sin^2{x}+\cos^2{x}=1\), it results in: \[(1-\sin^2{x})(1-\cos y)=\cos x\] Now, we want to find \(\cos{x}sec(y/2)\). Notice that \(\sec(y/2)=\frac{2}{1+\cos y}\). Thus, \[\cos{x}sec(y/2)=\frac{2\cos{x}}{1+\cos{y}}\] And by plugging in the previously obtained expression for \(\cos{x}\): \[\cos{x}sec(y/2)=\frac{2(1-\sin^2{x})(1-\cos y)}{(1+\cos y)^2}\] From the options provided, we can see that (B) \(±(1/\sqrt{2})\) is an option that satisfies the given equation. Thus, the answer is: \[\cos{x}sec(y/2)=±\frac{1}{\sqrt{2}}\]