A) sec x
B) sin x
C) tan x
D) cosec x
Correct Answer: A
Solution :
\[\frac{d}{dx}{{\cosh }^{-1}}(\sec x)\]\[=\frac{1}{\sqrt{{{\sec }^{2}}x-1}}\frac{d}{dx}\sec x\] \[=\frac{1}{\tan x}.\sec x.\tan x=\sec x\].
You need to login to perform this action.
You will be redirected in
3 sec
