Question

If $\cos (\theta -\alpha ),\cos \theta ,\cos (\theta +\alpha )$ are in $H.P.$, then $\cos \theta \sec \frac{\alpha }{2}=$

• A. $\pm \sqrt{2}$
• B. $\pm \sqrt{3}$
• C. $\pm \frac{1}{\sqrt{2}}$
• D. $Noneofthese$

Answer 1

The correct option is A $\pm \sqrt{2}$

As we know that

If $a,b,c$ are in $H.P$ then $b=\frac{2ac}{a+c}$

Given $\cos (\theta -\alpha ),\cos \theta ,\cos (\theta +\alpha )$ are in $H.P$

So $⟹\cos \theta =\frac{2\cos (\theta -\alpha )\cos (\theta +\alpha )}{\cos (\theta -\alpha )+\cos (\theta +\alpha )}$

$⟹\cos \theta =\frac{2({\cos }^2\theta -{\sin }^2\alpha )}{2\cos \theta \cos \alpha }$

$⟹{\cos }^2\theta \cos \alpha ={\cos }^2\theta -{\sin }^2\alpha$

$⟹{\cos }^2\theta =\frac{{\sin }^2\alpha }{1-\cos \alpha }=\frac{4{\sin }^2\frac{\alpha }{2}{\cos }^2\frac{\alpha }{2}}{2{\sin }^2\frac{\alpha }{2}}$

$⟹{\cos }^2\theta {\text{sec}}^2\frac{\alpha }{2}=2$

$⟹\cos \theta \text{sec}\frac{\alpha }{2}=\pm \sqrt{2}$