If x = sec θ - cos θ , y = sec n θ - cos n θ then d y d x =

Given:

$$\begin{gathered} x=\sec \theta -\cos \theta \\ y={\sec }^n\theta -{\cos }^n\theta \end{gathered}$$

Then, we have

$\frac{dx}{d\theta }=\sec \theta \tan \theta +\sin \theta ...\left(i\right)$

$\frac{dy}{d\theta }=n{\sec }^{n-1}\theta \left(\sec \theta \tan \theta \right)-n{\cos }^{n-1}\theta \left(-\sin \theta \right)$

$\Rightarrow \frac{dy}{d\theta }=n{\sec }^n\theta \tan \theta +n{\cos }^{n-1}\theta \left(\sin \theta \right)$

Now,

$\frac{dy}{dx}=\left\{\frac{{\displaystyle \frac{dy}{d\theta }}}{{\displaystyle \frac{dx}{d\theta }}}\right\}$

       $=\frac{n{\sec }^n\theta \tan \theta +n{\cos }^{n-1}\theta \left(\sin \theta \right)}{\sec \theta \tan \theta +\sin \theta }$

      $=n\left[\frac{{\displaystyle \frac{{\sec }^n\theta }{\cos \theta }}+{\cos }^{n-1}\theta }{{\displaystyle \frac{\sec \theta }{\cos \theta }}+1}\right]$

     $=n\left[\frac{{\displaystyle \frac{{\sec }^n\theta }{\cos \theta }+\frac{{\displaystyle {\cos }^n\theta }}{\cos \theta }}}{{\displaystyle \frac{\sec \theta }{\cos \theta }+1}}\right]$

  $=n\left[\frac{{\displaystyle {\sec }^n\theta +{\cos }^n\theta }}{{\displaystyle \sec \theta +\cos \theta }}\right]$

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

       $=n\left[\sqrt{\frac{y^2+4}{x^2+4}}\right]$.