Explanation

(a)
Given, $\cos{(\theta - \alpha),\cos{\theta\ and\cos(\theta + \alpha)}}$ are in HP
$\Rightarrow \ \frac{1}{COS\ (\theta - \alpha)},\frac{1}{COS\ \theta},\frac{1}{COS\ (\theta + \alpha)}$ will be in AP
$\Rightarrow \frac{2}{\cos\theta} = \frac{1}{\cos(\theta - \alpha)} + \frac{1}{\cos(\theta + \alpha)}$
$= \frac{\cos{(\alpha + \theta) + \cos{(\theta - \alpha)}}}{\cos^{2}{\theta - \sin^{2}\alpha}}$
$\Rightarrow \frac{2}{\cos\theta} = \frac{2\cos{\theta\cos\alpha}}{\cos^{2}{\theta - \sin^{2}\alpha}}$
$\Rightarrow \cos^{2}{\theta - \sin^{2}{\alpha = \cos^{2}{\theta\cos\alpha}}}$
$\Rightarrow \cos^{2}{\theta(1 - \cos{\alpha) = \sin^{2}\alpha}}$
$\Rightarrow \cos^{2}{\theta\left( 2\sin^{2}\frac{\alpha}{2} \right) = 4\sin^{2}{\frac{\alpha}{2}\cos^{2}\frac{\alpha}{2}}}$
$\Rightarrow \cos^{2}{\theta\sec^{2}{\frac{\alpha}{2} = 2\ \Rightarrow \cos{\theta\sec{\frac{\alpha}{2} = \pm \sqrt{2}}}}}$