A) \[\sin (\alpha +\beta )\sin (\alpha -\beta )\]
B) \[\cos (\alpha +\beta )\cos (\alpha -\beta )\]
C) \[\sin (\alpha -\beta )\cos (\alpha +\beta )\]
D) \[\sin (\alpha +\beta )\cos (\alpha -\beta )\]
Correct Answer: D
Solution :
\[{{\cos }^{2}}\left( \frac{\pi }{4}-\beta \right)-{{\sin }^{2}}\left( \alpha -\frac{\pi }{4} \right)\] \[=\cos \,\left( \frac{\pi }{4}-\beta +\alpha -\frac{\pi }{4} \right)\,\cos \,\left( \frac{\pi }{4}-\beta -\alpha +\frac{\pi }{4} \right)\,\] \[=\cos (\alpha -\beta )\cos \left( \frac{\pi }{2}-\overline{\alpha +\beta } \right)=\cos (\alpha -\beta )\sin (\alpha +\beta )\].
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