If x = sec θ - cos θ and y = sec n θ - cos n θ , then x 2 + 4 d y d x 2 =

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

$\Rightarrow \frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{\theta }}=\sec \theta \tan \theta +\sin \theta$

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

$\Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{\theta }}=n{\sec }^n\theta \tan \theta +n{\cos }^{n-1}\theta \sin \theta$

$\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\mathit{=}\frac{{\displaystyle \frac{dy}{d\theta }}}{{\displaystyle \frac{dx}{d\theta }}}=\frac{n\left({\sec }^n\theta \tan \theta +{\cos }^{n-1}\theta \sin \theta \right)}{\sec \theta \tan \theta +\sin \theta }$

$\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{n\left[{\displaystyle \frac{1}{{\cos }^{n+1}\theta }}+{\cos }^{n-1}\theta \right]}{{\displaystyle \frac{1}{{\cos }^2\theta }}+1}$

$\left(x^2+4\right){\left(\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\right)}^2=\left[{\left(\frac{1-{\cos }^2\theta }{\cos \theta }\right)}^2+4\right]\left[\frac{n^2{\left(1+{\cos }^{2n}\theta \right)}^2}{{\left(1+{\cos }^2\theta \right)}^2{\cos }^{2n-2}\theta }\right]$

$=n^2\left[\frac{{\left(1+{\cos }^{2n}\theta \right)}^2}{{\cos }^{2n}\theta }\right]$

$=n^2\left[\frac{1+co{s}^{4n}\theta +2{\cos }^{2n}\theta }{{\cos }^{2n}\theta }\right]$

$=n^2\left[\frac{{\left(1-{\cos }^{2n}\theta \right)}^2+4{\left({\cos }^n\theta \right)}^2}{{\left({\cos }^n\theta \right)}^2}\right]$

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

$=n^2\left[y^2+4\right]$