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