Question

If $y={\tan }^{-1}(\sec x-\tan x)$, then $\frac{dy}{dx}=$

• A. $2$
• B. $-2$
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$

Answer 1

Answer: (D)

$y={\tan }^{-1}\left(\sec x-\tan x\right)$

$\Rightarrow \frac{dy}{dx}=\frac{1.\left(\sec x\times \tan x-{\sec }^{2}x\right)}{1+{\left(\sec x-\tan x\right)}^2}$

$=\frac{\sec x\left(\tan x-\sec x\right)}{1+{\sec }^{2}x+{\tan }^{2}x-2\sec x\tan x}$

$=\frac{\sec x\left(\tan x-\sec x\right)}{2{\sec }^{2}x-2\sec x\tan x}$

$=\frac{\sec x\left(\tan x-\sec x\right)}{-2\sec x\left(\tan x-\sec x\right)}=\frac{-1}{2}$