Question

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

• A. $-\sqrt{2}$
• B. $\sqrt{2}$
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$

Answer 1

Answer: (A), (B)

The correct options are
A $-\sqrt{2}$
B $\sqrt{2}$

As $\cos (\theta -\alpha ),\cos \theta ,\cos (\theta +\alpha )$ In H.P
$\cos \theta =\frac{2\cos (\theta -\alpha )\cos (\theta +\alpha )}{\cos (\theta -\alpha )+\cos (\theta +\alpha )}=\frac{2({\cos }^2\theta -{\sin }^2\alpha )}{2\cos \theta \cos \alpha }\Rightarrow {\cos }^2\theta \cos \alpha ={\cos }^2\theta -{\sin }^2\alpha \Rightarrow (1-\cos \alpha ){\cos }^2\theta ={\sin }^2\alpha \Rightarrow {\cos }^2\theta =\frac{{\sin }^2\alpha }{(1-\cos \alpha )}=\frac{4{\sin }^2\left(\frac{\alpha }{2}\right){\cos }^2\left(\frac{\alpha }{2}\right)}{2{\sin }^2\left(\frac{\alpha }{2}\right)}\Rightarrow {\cos }^2\theta =2{\cos }^2\left(\frac{\alpha }{2}\right)\Rightarrow \cos \theta \sec \left(\frac{\alpha }{2}\right)=\pm \sqrt{2}$