Suppose theta and phi (ne 0) are such | vedclass

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Question

Since, $\sec \,(	heta  - \phi ),\,\,\sec \,	heta $ and $\sec \,(	heta  + \phi )$ are in $A.P.,$

$	herefore \,2\,\sec \,	he

Vedclass · Accepted Answer

$ \pm \sqrt 2 $

Vedclass · Answer

Since, $\sec \,(	heta  - \phi ),\,\,\sec \,	heta $ and $\sec \,(	heta  + \phi )$ are in $A.P.,$

$	herefore \,2\,\sec \,	heta \, = \,\sec \,(	heta  - \phi ) + \sec \,(	heta  + \phi )$

$ \Rightarrow \frac{2}{{\cos 	heta \,}} = \frac{{\cos \,(	heta  + \phi ) + \,\cos \,(	heta  - \phi )}}{{\cos \,(	heta  - \phi )\,\cos \,(	heta  + \phi )}}$

$ \Rightarrow \,2({\cos ^2}	heta  - {\sin ^2} \phi  )\, = \,\cos \,	heta \,[2\,\cos 	heta \,\cos \phi ]$

$ \Rightarrow \,{\cos ^2}	heta \,(1 - \cos \phi )\, = \,{\sin ^2}\phi \, = \,1 - {\cos ^2}\phi $

$ \Rightarrow \,{\cos ^2}	heta \, = 1 + \cos \phi \, = 2{\cos ^2}\frac{\phi }{2}$

$	herefore \,\,\cos \,	heta \, = \, \pm \,\sqrt 2 \cos \frac{\phi }{2}$

But given $\cos \,	heta \, = k\cos \frac{\phi }{2}$

$	herefore \,\,k = \,\, \pm \,\sqrt 2 $

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

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