If $x=\sec \theta -\cos \theta$ and $y={\sec }^n\theta -{\cos }^n\theta$ and $(x^2+4){\left(\frac{dy}{dx}\right)}^2=K(y^2+4)$ then k =___

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

$=Tan\theta (\sec \theta +\cos \theta )$

$\frac{dy}{d\theta }=n{\sec }^{n-1}\theta .Sec\theta Tan\theta -n{\cos }^{n-1}\theta (-\sin \theta )$

$=nTan\theta (Se{c}^n\theta +{\cos }^n\theta )$

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

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

$=\frac{n^2\left[{({\sec }^n\theta -{\cos }^n\theta )}^2+4\right]}{{(\sec \theta -\cos \theta )}^2+4}=\frac{n^2(y^2+4)}{x^2+4}$