Question

$\frac{d}{dx}{\cosh }^{-1}(\sec x)$ is equal to

• A. $sinx$
• B. $secx$
• C. $tanx$
• D. $cosecx$

Answer 1

Answer: (B)

$\frac{d}{dx}{\cosh }^{-1}(\sec x)$
$=\frac{1}{\sqrt{{\sec }^{2}x-1}}.\sec x\tan x$
$\left[\because \frac{d}{dx}{\cosh }^{-1}x=\frac{1}{\sqrt{x^2-1}}\right]$
$=\frac{\sec x.\tan x}{\sqrt{{\tan }^{2}x}}$
$=\frac{\sec x.\tan x}{\tan x}$ $=\sec x$