A) $ \sec A $
B) $ 2\sec A $
C) $ \sec 2A $
D) $ 2\sec 2A $
Correct Answer: D
$ (\sec 2A+1){{\sec }^{2}}A=( \frac{1+{{\tan }^{2}}A}{1-{{\tan }^{2}}A}+1 ),(1+{{\tan }^{2}}A) $ $ =\frac{2,(1+{{\tan }^{2}}A)}{1-{{\tan }^{2}}A}=2\sec 2A. $
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