A) \[4{{\cos }^{2}}\alpha \]
B) \[4{{\cos }^{2}}\alpha -1\]
C) \[2{{\cos }^{2}}\alpha -2{{\sin }^{2}}\alpha \]
D) \[{{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha \]
Correct Answer: C
Solution :
Given, \[x+\frac{1}{x}=2\cos \alpha \] On squaring both sides, we get \[{{x}^{2}}+\frac{1}{{{x}^{2}}}+2=4{{\cos }^{2}}\alpha \] \[\Rightarrow \]\[{{x}^{2}}+\frac{1}{{{x}^{2}}}=2\,\,(2{{\cos }^{2}}\alpha -1)=2\,\,(2{{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha -{{\cos }^{2}}\alpha )\] \[=2{{\cos }^{2}}\alpha -2{{\sin }^{2}}\alpha \]
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