[AISSE 1982]
A) $ \cos x $
B) $ \sec x $
C) $ \tan x $
D) $ \cot x $
Correct Answer: B
$ \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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