Question

If $\sec (\varphi -\alpha ),\sec \varphi ,\sec (\varphi +\alpha )$ are in A.P., prove that $\cos \varphi =\sqrt{2}\cos \left(\alpha /2\right)$.

Answer 1

Since $\sec (\varphi -\alpha ),\sec \varphi ,\sec (\varphi +\alpha )$ are in A.P
$\therefore 2\sec \varphi =\sec (\varphi -\alpha )+\sec (\varphi +\alpha )$
or $\frac{2}{\cos \varphi }=\frac{\cos (\varphi +\alpha )+\cos (\varphi -\alpha )}{\cos (\varphi +\alpha )-\cos (\varphi -\alpha )}$
or $2.({\cos }^2\varphi -{\sin }^2\varphi )=\cos \varphi \left[2\cos \varphi \cos \alpha \right]$
${\cos }^2\varphi =1+\cos \alpha =2{\cos }^2\left(\alpha /2\right)$
$\therefore \cos \varphi =\sqrt{2}\cos \left(\alpha /2\right)$