A) \[\pm \sqrt{2}\]
B) \[\pm 1\]
C) \[\pm \frac{1}{\sqrt{2}}\]
D) \[\pm 2\]
Correct Answer: A
Solution :
Since, \[\sec (\theta -\phi ),\sec \theta \]and \[\sec (\theta +\phi )\]are in A.P., \[\therefore \]\[2\sec \theta =\sec (\theta -\phi )+\sec (\theta +\phi )\] \[\Rightarrow \]\[\frac{2}{\cos \theta }=\frac{\cos (\theta +\phi )+\cos (gq-\phi )}{\cos (\theta -\phi )\cos (\theta +\phi )}\] \[\Rightarrow \]\[^{2}\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\phi \right)=\cos \theta [2\cos \theta \cos \phi ]\] \[\Rightarrow \]\[{{\cos }^{2}}\theta \left( 1-\cos \,\phi \right)={{\sin }^{2}}\phi =1-{{\cos }^{2}}\phi \] \[\Rightarrow \]\[{{\cos }^{2}}\theta =1+\cos \phi =2{{\cos }^{2}}\frac{\phi }{2}\] \[\therefore \]\[\cos \theta =\sqrt{2}\cos \frac{\phi }{2}\] But given \[\cos \theta =k\cos \frac{\phi }{2}\] \[\therefore \]\[k=\sqrt{2}\]
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