A) 1
B) 1/2
C) \[\cos x\]
D) \[\sec x\]
Correct Answer: B
Solution :
\[\frac{d}{dx}{{\tan }^{-1}}(\sec x+\tan x)=\frac{d}{dx}{{\tan }^{-1}}\left( \frac{1+\sin x}{\cos x} \right)\] \[=\frac{d}{dx}{{\tan }^{-1}}\left( \frac{\sin \left( \frac{x}{2} \right)+\cos \left( \frac{x}{2} \right)}{\cos \left( \frac{x}{2} \right)-\sin \left( \frac{x}{2} \right)} \right)=\frac{d}{dx}\left( \frac{\pi }{4}+\frac{x}{2} \right)=\frac{1}{2}\].
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