Question

If $y={\left(tan\right)}^{-1}\left(secx-tanx\right),$ then the value of $\left|2\left(\frac{dy}{dx}\right)\right|$ is

Answer 1

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
$\Rightarrow \frac{dy}{dx}=\frac{-1}{2}$
Therefore, $\left|2\left(\frac{dy}{dx}\right)\right|=1$.