If $\sec (\theta +\alpha )+\sec (\theta -\alpha )=2\sec \theta$  and $\cos \alpha \ne 1$  then show that  $\cos \theta =\pm \sqrt{2}\cos \left(\frac{\alpha }{2}\right)$

$\mathrm{Sec}(\theta +\alpha )+\mathrm{Sec}(\theta -\alpha )=2\mathrm{Sec}\theta$

$\Rightarrow \frac{1}{\mathrm{Cos}(\theta +\alpha )}+\frac{1}{\mathrm{Cos}(\theta -\alpha )}=\frac{2}{\mathrm{Cos}\theta }$

$\Rightarrow \frac{\mathrm{Cos}(\theta -\alpha )+\mathrm{Cos}(\theta +\alpha )}{\mathrm{Cos}(\theta +\alpha )\mathrm{Cos}(\theta -\alpha )}=\frac{2}{\mathrm{Cos}\theta }$

$\Rightarrow \frac{2\mathrm{Cos}\theta \mathrm{Co}s\alpha }{{\mathrm{Cos}}^2\theta -{\mathrm{Sin}}^2\alpha }=\frac{2}{\mathrm{Cos}\theta }$

$\Rightarrow {\mathrm{Cos}}^2\theta \mathrm{Cos}\alpha ={\mathrm{Cos}}^2\theta -{\mathrm{Sin}}^2\alpha$

$\Rightarrow {\mathrm{Sin}}^2\alpha ={\mathrm{Cos}}^2\theta -{\mathrm{Cos}}^2\theta \mathrm{Cos}\alpha$

$\Rightarrow {\mathrm{Sin}}^2\alpha ={\mathrm{Cos}}^2\theta (1-\mathrm{Cos}\alpha )$

$\Rightarrow 1-{\mathrm{Cos}}^2\alpha ={\mathrm{Cos}}^2\theta (1-\mathrm{Cos}\alpha )$

$\Rightarrow {\mathrm{Cos}}^2\theta =\frac{1-{\mathrm{Cos}}^2\alpha }{1-\mathrm{Cos}\alpha }=\frac{(1+\mathrm{Co}s\alpha )(1-\mathrm{Cos}\alpha )}{1-\mathrm{Cos}\alpha }$

$\Rightarrow {\mathrm{Cos}}^2\theta =1+\cos \alpha \Rightarrow {\mathrm{Cos}}^2\theta =2{\mathrm{Cos}}^2\frac{\alpha }{2}$

$\Rightarrow \mathrm{Cos}\theta =\pm \sqrt{2{\mathrm{Cos}}^2\alpha /2}=\pm \sqrt{2}\mathrm{Cos}\alpha /2$