A) \[\sin 2x\]
B) \[\cos 2x\]
C) \[\tan 2x\]
D) None of these
Correct Answer: D
Solution :
\[\frac{\sin x-\sin 3x}{{{\sin }^{2}}x-{{\cos }^{2}}x}=\frac{\sin 3x-\sin x}{{{\cos }^{2}}x-{{\sin }^{2}}x}\] \[=\frac{2\cos \frac{3x+x}{2}.\sin \frac{3x-x}{2}}{\cos 2x}=\frac{2\cos 2x.\sin x}{\cos 2x}\] 0\[=2\sin x\]
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