A) \[0\]
B) \[\cos \,\,2\theta \]
C) \[sin\,\,2\theta \]
D) \[\cos \,\theta \]
Correct Answer: A
Solution :
We know that, \[{{\cos }^{2}}\,(A)-{{\sin }^{2}}(B)\] \[=\cos (A+B)\,\cos \,(A-B)\] \[\therefore \] \[{{\cos }^{2}}\,\left( \frac{\pi }{4}+\theta \right)-{{\sin }^{2}}\left( \frac{\pi }{4}-\theta \right)\] \[=\cos \,\left( \frac{\pi }{4}+\theta +\frac{\pi }{4}-\theta \right)\,\,\cos \,\left( \frac{\pi }{4}+\theta -\frac{\pi }{4}+\theta \right)\] \[=\cos \,\left( \frac{\pi }{2} \right)\,\cos \,(2\,\theta )=0\]
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