A) \[-\sin 2x\]
B) \[2\sin 2x\]
C) \[2\cos 2x\]
D) \[-2\sin 2x\]
Correct Answer: D
Solution :
\[\frac{d}{dx}\left[ \frac{{{\cot }^{2}}x-1}{{{\cot }^{2}}x+1} \right]=\frac{d}{dx}\left[ \frac{{{\cos }^{2}}x-{{\sin }^{2}}x}{{{\cos }^{2}}x+{{\sin }^{2}}x} \right]\] \[=\frac{d}{dx}[\cos 2x]=-2\sin 2x\].You need to login to perform this action.
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