Search any question & find its solution
Question:
Answered & Verified by Expert
If $y=\tan ^{-1}(\sec x-\tan x)$, then $\frac{d y}{d x}$ is equal to
Options:
Solution:
1042 Upvotes
Verified Answer
The correct answer is:
$-\frac{1}{2}$
We have,
$y=\tan ^{-1}(\sec x-\tan x)$
$y=\tan ^{-1}\left[\frac{1-\sin x}{\cos x}\right]$
$\begin{aligned} &=\tan ^{-1}\left[\frac{1-\cos \left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)}\right] \\ &=\tan ^{-1}\left[\frac{2 \sin ^{2}\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2 \sin \left(\frac{\pi}{4}-\frac{x}{2}\right) \cos \left(\frac{\pi}{4}-\frac{x}{2}\right)}\right] \\ &=\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right\}=\frac{\pi}{4}-\frac{x}{2} \\ \frac{d y}{d x} &=-\frac{1}{2} \end{aligned}$
$y=\tan ^{-1}(\sec x-\tan x)$
$y=\tan ^{-1}\left[\frac{1-\sin x}{\cos x}\right]$
$\begin{aligned} &=\tan ^{-1}\left[\frac{1-\cos \left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)}\right] \\ &=\tan ^{-1}\left[\frac{2 \sin ^{2}\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2 \sin \left(\frac{\pi}{4}-\frac{x}{2}\right) \cos \left(\frac{\pi}{4}-\frac{x}{2}\right)}\right] \\ &=\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right\}=\frac{\pi}{4}-\frac{x}{2} \\ \frac{d y}{d x} &=-\frac{1}{2} \end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.