$\text{If }\sec (\mathrm{\theta }+\mathrm{\alpha }),\sec \mathrm{\theta },\sec (\mathrm{\theta }-\mathrm{\alpha })$ are in A.P. then $\cos \mathrm{\theta sec}\mathrm{\alpha }/2=$

Given that $\sec (\mathrm{\theta }+\mathrm{\alpha }),\sec \mathrm{\theta },\sec (\mathrm{\theta }-\mathrm{\alpha })\text{ are in A.P. }\Rightarrow 2\sec \mathrm{\theta }=\sec (\mathrm{\theta }+\mathrm{\alpha })+\sec (\mathrm{\theta }-\mathrm{\alpha })$
$\begin{array}{l}\Rightarrow \frac{2}{\cos \mathrm{\theta }}=\frac{1}{\cos (\mathrm{\theta }+\mathrm{\alpha })}+\frac{1}{\cos (\mathrm{\theta }-\mathrm{\alpha })}\\ \Rightarrow 2\cos (\mathrm{\theta }+\mathrm{\alpha })\cos (\mathrm{\theta }-\mathrm{\alpha })=\cos \mathrm{\theta }[\cos (\mathrm{\theta }+\mathrm{\alpha })+\cos (\mathrm{\theta }-\mathrm{\alpha }\left)\right]\\ \Rightarrow 2\left({\cos }^2\mathrm{\theta }-{\sin }^2\mathrm{\alpha }\right)=\cos \mathrm{\theta }\cdot 2\cos \mathrm{\theta cos}\mathrm{\alpha }\Rightarrow {\cos }^2\mathrm{\theta }(1-\cos \mathrm{\alpha }){\sin }^2\mathrm{\alpha }\\ \Rightarrow {\cos }^2\mathrm{\theta }\cdot 2{\sin }^2\frac{\mathrm{\alpha }}{2}=4{\sin }^2\mathrm{\alpha }/2{\cos }^2\mathrm{\alpha }/2\Rightarrow {\cos }^2\mathrm{\theta }\cdot {\sec }^2\mathrm{\alpha }/2=2\Rightarrow \cos \mathrm{\theta sec}\mathrm{\alpha }/2=\pm \sqrt{2}\end{array}$