A) \[\frac{2\cos x}{{{(1-\sin x)}^{2}}}\]
B) \[\frac{\cos x}{{{(1-\sin x)}^{2}}}\]
C) \[\frac{2\cos x}{1-\sin x}\]
D) None of these
Correct Answer: A
Solution :
\[\frac{d}{dx}\left( \frac{\sec x+\tan x}{\sec x-\tan x} \right)=\frac{d}{dx}\left( \frac{1+\sin x}{1-\sin x} \right)\].
You need to login to perform this action.
You will be redirected in
3 sec
