A) \[\frac{3}{2}\]
B) \[\frac{2}{3}\]
C) \[\frac{3+\sqrt{3}}{2}\]
D) \[\frac{2}{3+\sqrt{3}}\]
Correct Answer: A
Solution :
\[{{\cos }^{2}}\frac{\pi }{12}+{{\cos }^{2}}\frac{\pi }{4}+{{\cos }^{2}}\frac{5\pi }{12}\] \[=1-{{\sin }^{2}}\left( \frac{\pi }{12} \right)+{{\left( \frac{1}{\sqrt{2}} \right)}^{2}}+{{\cos }^{2}}\left( \frac{5\pi }{12} \right)\] \[=1+\frac{1}{2}+\left( {{\cos }^{2}}\frac{5\pi }{12}-{{\sin }^{2}}\frac{\pi }{12} \right)\] \[=\frac{3}{2}+\cos \left( \frac{5\pi }{12}+\frac{\pi }{12} \right)\cos \left( \frac{5\pi }{12}-\frac{\pi }{12} \right)=\frac{3}{2}+\cos \frac{\pi }{2}\cos \frac{\pi }{3}\] \[=\frac{3}{2}+0.\frac{1}{2}=\frac{3}{2}\].
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