Question

Suppose $\theta$ and $\varphi (\ne 0)$ are such that $\sec (\theta +\varphi ),\sec \theta$ and $\sec (\theta -\varphi )$ are in $A.P$. If $\cos \theta =k\cos \left(\frac{\varphi }{2}\right)$ for some $k$, then $k$ is equal to

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

Answer 1

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

$2\sec \theta =\sec (\theta +\varphi )+\sec (\theta -\varphi )$

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

$2({\cos }^2\theta -{\sin }^2\varphi )=2{\cos }^2\theta \cos \varphi$

${\cos }^2\theta (1-\cos \varphi )={\sin }^2\varphi$
${\cos }^2\theta (1-\cos \varphi )=1-{\cos }^2\varphi$

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

$\cos \theta =\pm \sqrt{2}\cos \frac{\varphi }{2}$

$⟹k=\pm \sqrt{2}$