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 } $

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

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