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?

Answer 1

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

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

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

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

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

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

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

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

$k=\sqrt{2}$