A) \[y=\sec x+\tan x+c\]
B) \[y=\sec x+\cot x+c\]
C) \[y=\sec x-\tan x+c\]
D) None of these
Correct Answer: A
Solution :
\[\frac{dy}{dx}=\sec x(\sec x+\tan x)\] Þ \[\frac{dy}{dx}={{\sec }^{2}}x+\sec x\tan x\] Now integrating both sides, we get \[y=\tan x+\sec x+c\].You need to login to perform this action.
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