A) \[sin\text{ }x\]
B) \[sec\text{ }x\]
C) \[tan\text{ }x\]
D) \[cosec\text{ }x\]
Correct Answer: B
Solution :
\[\frac{d}{dx}{{\cosh }^{-1}}(\sec x)\] \[=\frac{1}{\sqrt{{{\sec }^{2}}x-1}}.\sec x\tan x\] \[\left[ \because \frac{d}{dx}{{\cosh }^{-1}}x=\frac{1}{\sqrt{{{x}^{2}}-1}} \right]\] \[=\frac{\sec x.\tan x}{\sqrt{{{\tan }^{2}}x}}\] \[=\frac{\sec x.\tan x}{\tan x}\] \[=\sec x\]
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