If \[\cos \theta =\frac{1}{2}\left( x+\frac{1}{x} \right)\], then \[\frac{1}{2}\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)=\] [AMU 1998]

If \[\cos \theta =\frac{1}{2}\left( x+\frac{1}{x} \right)\], then \[\f

1.	If \[\cos \theta =\frac{1}{2}\left( x+\frac{1}{x} \right)\], then \[\frac{1}{2}\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)=\] [AMU 1998]
A.	\[\sin 2\theta \]
B.	\[\cos \,2\theta \]
C.	\[\tan \,2\theta \]
D.	\[\sec \,2\theta \]
Answer» C. \[\tan \,2\theta \]

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If \[\cos \theta =\frac{1}{2}\left( x+\frac{1}{x} \right)\], then \[\frac{1}{2}\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)=\] [AMU 1998]

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