A) $ {{\sec }^{2}}x $
B) $ -cose{c^{2}}x $
C) $ 2{{\sec }^{2}}\frac{x}{2} $
D) $ -2\csc^{2}\frac{x}{2} $
Correct Answer: B
$ \frac{d}{dx}\sqrt{\frac{1+\cos 2x}{1-\cos 2x}}=\frac{d}{dx}\cot x=-\csc^{2}x $
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