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

$x=\sec \theta -\cos \theta$

$\frac{dx}{d\theta }=\sec \theta .\tan \theta +\sin \theta$

$y={\sec }^n\theta -{\cos }^n\theta$

$\frac{dy}{d\theta }=n.{\sec }^{n-1}\theta .\sec \theta .\tan \theta +n{\cos }^{n-1}\theta .\sin \theta$

$\frac{dy}{dx}=\frac{n.{\sec }^n\theta .\tan \theta +n.{\cos }^{n-1}\theta .\sin \theta }{\sec \theta .\tan \theta +\sin \theta }$

$=\frac{n.{\sec }^n\theta .\frac{\sin \theta }{\cos \theta }+n.{\cos }^{n-1}\theta .\sin \theta }{\frac{1}{\cos \theta .}\frac{\sin \theta }{\cos \theta }+\sin \theta }$

$=\frac{n.{\sec }^n\theta .\sin \theta +n.{\cos }^n\theta .\sin \theta }{\tan \theta +\sin \theta .\cos \theta }$

$=\frac{n\sin \theta ({\sec }^n\theta +{\cos }^n\theta )}{\sin \theta (\sec \theta +\cos \theta )}$

${\left(\frac{dy}{dx}\right)}^2=\frac{n^2{({\sec }^n\theta +{\cos }^n\theta )}^2}{{(\sec \theta +\cos \theta )}^2}$

$=\frac{n^2(y^2+4)}{x^2+4}$

$[\therefore y^2+4={\sec }^{2n}\theta +{\cos }^{2n}\theta -2{\sec }^n\theta {\cos }^n\theta +4{\sec }^n\theta .{\cos }^n\theta ={({\sec }^n\theta +{\cos }^n\theta )}^2]$

$[x^2+4={\sec }^2\theta +{\cos }^2\theta -2\sec \theta \cos \theta +4\sec \theta .\cos \theta ={(\sec \theta +\cos \theta )}^2]$