Question

If $\cos \left(\theta -\alpha \right),\cos \theta$ and $\cos \left(\theta +\alpha \right)$ are in H.P., then $\cos \theta \sec \frac{\alpha }{2}$ is equal to

• A. $\pm \sqrt{2}$
• B. $\pm \sqrt{3}$
• C. $-3$
• D. $-2$

Answer 1

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

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

$\frac{1}{\cos (\theta -\alpha )},\frac{1}{\cos \theta },\frac{1}{\cos (\theta +\alpha )}-A.P$

$\frac{1}{\cos (\theta -\alpha )}+\frac{1}{\cos (\theta +\alpha )}=\frac{2}{\cos \theta }$

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

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

${\cos }^2\theta =\cos \alpha +1$

${\cos }^2\theta =2{\cos }^2\left(\frac{\alpha }{2}\right)$

${\cos }^2\theta {\sec }^2\left(\frac{\alpha }{2}\right)=2$

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

$A$ is correct