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Suppose $\theta $ and $\phi (\ne 0)$ are such that $sec\,(\theta + \phi ),$ $sec\,\theta $ and $sec\,(\theta - \phi )$ are in $A.P.$ If $cos\,\theta = k\,cos\,( \frac {\phi }{2})$ for some $k,$ then $k$ is equal to
$ \pm \sqrt 2 $
$ \pm 1 $
$ \pm \frac{1}{{\sqrt 2 }}$
$ \pm 2 $
Solution
Since, $\sec \,(\theta – \phi ),\,\,\sec \,\theta $ and $\sec \,(\theta + \phi )$ are in $A.P.,$
$\therefore \,2\,\sec \,\theta \, = \,\sec \,(\theta – \phi ) + \sec \,(\theta + \phi )$
$ \Rightarrow \frac{2}{{\cos \theta \,}} = \frac{{\cos \,(\theta + \phi ) + \,\cos \,(\theta – \phi )}}{{\cos \,(\theta – \phi )\,\cos \,(\theta + \phi )}}$
$ \Rightarrow \,2({\cos ^2}\theta – {\sin ^2} \phi )\, = \,\cos \,\theta \,[2\,\cos \theta \,\cos \phi ]$
$ \Rightarrow \,{\cos ^2}\theta \,(1 – \cos \phi )\, = \,{\sin ^2}\phi \, = \,1 – {\cos ^2}\phi $
$ \Rightarrow \,{\cos ^2}\theta \, = 1 + \cos \phi \, = 2{\cos ^2}\frac{\phi }{2}$
$\therefore \,\,\cos \,\theta \, = \, \pm \,\sqrt 2 \cos \frac{\phi }{2}$
But given $\cos \,\theta \, = k\cos \frac{\phi }{2}$
$\therefore \,\,k = \,\, \pm \,\sqrt 2 $
