If $x=\sec \theta -\cos \theta ,$ $y={\sec }^n\theta -{\cos }^n\theta$ then ${\left[\frac{dy}{dx}\right]}^2=$

$\frac{dy}{dx}=\frac{dy}{d\theta }/\frac{dx}{d\theta }$
$\frac{n\left({\sec }^n\theta Tan\theta \right)+n\left({\cos }^{n-1}\theta \right)\sin \theta }{\sec \theta Tan\theta +\sin \theta }$
$\frac{n\left({\sec }^n\theta Tan\theta \right)+n\left(\frac{{\cos }^n\theta }{\cos \theta }\right)\sin \theta )}{\sec \theta Tan\theta +\cos \theta \left(\frac{\sin \theta }{\cos \theta }\right)}$
$\frac{n\left(Tan\theta \right)({\sec }^n\theta +{\cos }^n\theta )}{Tan\theta (\sec \theta +\cos \theta )}$
$\frac{dy}{dx}=\frac{n({\sec }^n\theta +{\cos }^n\theta )}{\sec \theta +\cos \theta }$
${\left(\frac{dy}{dx}\right)}^2=\frac{n^2({({\sec }^n\theta -{\cos }^n\theta )}^2+4)}{{(\sec \theta -\cos \theta )}^2+4}$
$=\frac{x^2(y^2+4)}{x^2+4}$