\[\frac{d}{dx}{{\tan }^{-1}}(\sec x+\tan x)=\] [AISSE 1985, 87; DSSE 1982, 84]

\[\frac{d}{dx}{{\tan }^{-1}}(\sec x+\tan x)=\] [AISSE 1985, 87; DSSE

1.	\[\frac{d}{dx}{{\tan }^{-1}}(\sec x+\tan x)=\] [AISSE 1985, 87; DSSE 1982, 84]
A.	1
B.	1/2
C.	\[\cos x\]
D.	\[\sec x\]
Answer» C. \[\cos x\]

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\[\frac{d}{dx}{{\tan }^{-1}}(\sec x+\tan x)=\] [AISSE 1985, 87; DSSE 1982, 84]

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