Question

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

Answer 1

Answer: -0.5

$y={\left(tan\right)}^{-1}\left(secx-tanx\right)$
$y={\left(tan\right)}^{-1}\left(\frac{1-sinx}{cosx}\right)$
$\Rightarrow y=(tan{)}^{-1}(\frac{(cos\frac{x}{2}-sin\frac{x}{2}{)}^2}{(cos{)}^2\frac{x}{2}-(sin{)}^2\frac{x}{2})})$
$\Rightarrow y={\left(tan\right)}^{-1}\left(\frac{cos\frac{x}{2}-sin\frac{x}{2}}{cos\frac{x}{2}+sin\frac{x}{2}}\right)$
$\Rightarrow y={\left(tan\right)}^{-1}\left(\frac{1-tan\frac{x}{2}}{1+tan\frac{x}{2}}\right)$
$\Rightarrow y=(tan{)}^{-1}(tan(\frac{\pi }{4}-\frac{x}{2}))$
$\Rightarrow y=\frac{\pi }{4}-\frac{x}{2}$
Differentiating with respect to $x,$ we get
$\Rightarrow \frac{dy}{dx}=\frac{-1}{2}=-0.5$