Question

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

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

Answer 1

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

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

$={\tan }^{-1}(\frac{1}{\cos x}-\frac{\sin x}{\cos x})$

$={\tan }^{-1}\left(\frac{1-\sin x}{\cos x}\right)$

$={\tan }^{-1}\left[\frac{1-\cos (\pi /2-x)}{\sin (\pi /2-x)}\right]$

as $\cos x=1-2{\sin }^2\frac{x}{2}$ & $\sin x=2\sin \frac{x}{2}\cos \frac{x}{2}$

$y={\tan }^{-1}\left[\frac{1-1+2\sin (\pi /4-x/2)}{2\sin (\pi /2-x/2)\cos (\pi /2-x/2)}\right]$

$={\tan }^{-1}\left[\frac{{\sin }^2(\pi /4-x/2)}{\sin (\pi /4-x/2)\cos (\pi /4-x/2)}\right]$

${\tan }^{-1}\left(\frac{\sin (\pi /4-x/2)}{\cos (\pi /4-x/2)}\right)$

$={\tan }^{-1}\left(\tan (\frac{\pi }{4}-\frac{x}{2})\right)$

$y=\frac{\pi }{4}-\frac{x}{2}$

$\frac{dy}{dx}=\frac{d}{dx}(\frac{\pi }{4}-\frac{x}{2})$

$=0-\frac{1}{2}$

$\therefore \frac{dy}{dx}=\frac{-1}{2}$, op[tion $B$ is correct