If $\cos \left(\theta -\varphi \right),\cos \theta$ and $\cos \left(\theta +\varphi \right)$ are in $H.P$  then ${\left(\cos \theta \sec \left(\frac{\varphi }{2}\right)\right)}^2=$

$\Rightarrow \cos \theta =\frac{2\cos \left[\theta -\varphi \right].\cos \left[\theta +\varphi \right]}{\cos \left[\theta +\varphi \right]+\cos \left[\theta -\varphi \right]}$

$\Rightarrow \cos \theta =\frac{2\left[{\cos }^2\theta -{\sin }^2\varphi \right]}{2\cos \theta .\cos \varphi }$

$\Rightarrow \cos \theta =\frac{{\cos }^2\theta -{\sin }^2\varphi }{\cos \theta .\cos \varphi }$

$\Rightarrow {\cos }^2\theta .\cos \varphi ={\cos }^2\theta -{\sin }^2\varphi$

$\Rightarrow {\sin }^2\varphi ={\cos }^2\theta -{\cos }^2\theta .\cos \varphi$

$\Rightarrow 1-{\cos }^2\varphi ={\cos }^2\theta \left[1-\cos \varphi \right]$

$\left[1-\cos \theta \right]\left[1+\cos \varphi \right]={\cos }^2\theta \left[1-\cos \varphi \right]$

$\Rightarrow 1+\cos \varphi ={\cos }^2\theta$

$\Rightarrow 2{\cos }^2\frac{\varphi }{2}={\cos }^2\theta$

$\Rightarrow 2={\cos }^2\theta .{\sec }^2\left[\frac{\varphi }{2}\right]$

$\Rightarrow \pm \sqrt{2}=\cos \theta .\sec \left[\frac{\varphi }{2}\right]$

${\left(\cos \theta \sec \left(\frac{\varphi }{2}\right)\right)}^2={\left(\pm \sqrt{2}\right)}^2=2$