Question

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

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

Answer 1

The correct option is D $-\frac{1}{2}$

Given that $y={\tan }^{-1}(\sec x-\tan x)$
$y={\tan }^{-1}\left(\frac{1-\sin x}{\cos x}\right)$
$y={\tan }^{-1}\left(\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right)$
$y={\tan }^{-1}\left[\tan (\frac{\pi }{4}-\frac{\pi }{2})\right]$
$y=\frac{\pi }{4}-\frac{x}{2}$
$\Rightarrow \frac{dy}{dx}=0-\frac{1}{2}$
$\Rightarrow \frac{dy}{dx}=-\frac{1}{2}$