A) \[\cos x\]
B) \[\sec x\]
C) \[\tan x\]
D) \[\cot x\]
Correct Answer: B
Solution :
\[\frac{d}{dx}\{\log (\sec x+\tan x)\}=\frac{\sec x\tan x+{{\sec }^{2}}x}{\sec x+\tan x}=\sec x\].
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