A) 3/2
B) 1
C) 1/2
D) 0
Correct Answer: A
Solution :
\[{{\cos }^{2}}\alpha +{{\cos }^{2}}(\alpha +{{120}^{o}})+{{\cos }^{2}}(\alpha -{{120}^{o}})\] \[={{\cos }^{2}}\alpha +{{\left\{ \cos \,(\alpha +{{120}^{o}})+\cos \,(\alpha -{{120}^{o}}) \right\}}^{2}}\]\[-2\,\cos \,(\alpha +{{120}^{o}})\,\cos \,(\alpha -{{120}^{o}})\] \[={{\cos }^{2}}\alpha +{{\left\{ \,2\,\cos \,\,\alpha \,\cos \,{{120}^{o}} \right\}}^{2}}-2\,\left\{ {{\cos }^{2}}\alpha -{{\sin }^{2}}\,{{120}^{o}} \right\}\] \[={{\cos }^{2}}\alpha +{{\cos }^{2}}\alpha -2\,{{\cos }^{2}}\alpha +2\,{{\sin }^{2}}\,{{120}^{o}}\] \[=2{{\sin }^{2}}{{120}^{o}}=2\times \frac{3}{4}=\frac{3}{2}\].
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