A) \[\frac{1}{2}\cos 2\theta \]
B) 0
C) \[-\frac{1}{2}\cos 2\,\theta \]
D) \[\frac{1}{2}\]
Correct Answer: A
Solution :
\[{{\cos }^{2}}\left( \frac{\pi }{6}+\theta \right)-{{\sin }^{2}}\left( \frac{\pi }{6}-\theta \right)\] \[=\cos \left( \frac{\pi }{6}+\theta +\frac{\pi }{6}-\theta \right)\cos \left( \frac{\pi }{6}+\theta -\frac{\pi }{6}+\theta \right)\]\[[\because {{\cos }^{2}}A-{{\sin }^{2}}B=\cos (A+B)\cos (A-B)]\] \[=\cos \frac{2\pi }{6}\cos 2\theta =\frac{1}{2}\cos 2\theta \].
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