A) \[\pi \]
B) \[2\pi \]
C) \[\frac{\pi }{2}\]
D) None of these
Correct Answer: C
Solution :
Given, \[f(x)=\frac{2\sin 8x\cos x-2\sin 6x\cos 3x}{2\cos 2x\cos x-2\sin 3x\sin 4x}\] \[=\frac{(\sin 9x+\sin 7x)-(\sin 9x+\sin 3x)}{(\cos 3x+\cos x)+(\cos 7x-\cos x)}\] \[=\frac{\sin 7x-\sin 3x}{\cos 7x+\cos 3x}=\frac{2\sin 2x\cos 5x}{2\cos 5x\cos 2x}\] \[=\tan 2x\] \[\therefore \]Period of\[f(x)=\frac{\pi }{2}\]
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