A) \[\sec A\]
B) \[2\sec A\]
C) \[\sec 2A\]
D) \[2\sec 2A\]
Correct Answer: D
Solution :
\[(\sec 2A+1){{\sec }^{2}}A=\left( \frac{1+{{\tan }^{2}}A}{1-{{\tan }^{2}}A}+1 \right)\,(1+{{\tan }^{2}}A)\] \[=\frac{2\,(1+{{\tan }^{2}}A)}{1-{{\tan }^{2}}A}=2\sec 2A.\]
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