A) \[\sin 2\theta \]
B) \[\cos \,2\theta \]
C) \[\tan \,2\theta \]
D) \[\sec \,2\theta \]
Correct Answer: B
Solution :
Given that \[\cos \theta =\frac{1}{2}\,\left( x+\frac{1}{x} \right)\,\,\Rightarrow \,x+\frac{1}{x}=2\,\cos \theta \] We know that \[{{x}^{2}}+\frac{1}{{{x}^{2}}}={{\left( x+\frac{1}{x} \right)}^{2}}-2\] \[={{(2\cos \theta )}^{2}}-2=4\,{{\cos }^{2}}\theta -2=2\,\cos \,\,2\theta \] \[\therefore \,\,\frac{1}{2}\,\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)=\frac{1}{2}\times 2\,\cos \,2\theta =\cos \,2\theta \]
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