Question

If $y=\left(t a n\right)^{- 1}\left(s e c x - t a n x\right),$ then the value of $\left|2 \left(\frac{d y}{d x}\right)\right|$ is

Answer 1

$y=\tan ^{-1}(\sec x-\tan x)$
$y=\tan ^{-1}\left(\frac{1-\sin x}{\cos x}\right) $
$\Rightarrow y=\tan ^{-1}\left(\frac{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^{2}}{\left(\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}\right)}\right)$
$\Rightarrow y=\tan ^{-1}\left(\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right) $
$\Rightarrow y=\tan ^{-1}\left(\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right) $
$\Rightarrow y=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right)$
$\Rightarrow y=\frac{\pi}{4}-\frac{x}{2}$
Differentiating with respect to $x$, we get
$\Rightarrow \frac{d y}{d x}=\frac{-1}{2}$
Therefore, $\left|2\left(\frac{d y}{d x}\right)\right|=1$.