If \(x=\sec \theta-\cos \theta\) and \(y=\sec ^n \theta-\cos ^n \theta\), then \(\left(\frac{d y}{d x}\right)^2\) is equal to

\(\begin{aligned} & =\left(\frac{d y}{d x}\right)^2=\left(\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}\right)^2=\left\{\frac{\frac{d\left(\sec ^n \theta-\cos ^n \theta\right)}{d \theta}}{\frac{d(\sec \theta-\cos \theta)}{d \theta}}\right\}^2 \\ & =\left\{\frac{n \sec ^{\mathrm{n}-1} \theta \cdot \sec \theta \cdot \tan \theta-\mathrm{n} \cos ^{\mathrm{n}-1} \theta \cdot(-\sin \theta)}{\sec \theta \cdot \tan \theta-(-\sin \theta)}\right\}^2 \\ & =\left\{\frac{\mathrm{n} \sec ^{\mathrm{n}} \theta \cdot \tan \theta+\mathrm{n} \cos ^{\mathrm{n}} \theta \cdot \tan \theta}{\sec \theta \cdot \tan \theta+\cos \theta \cdot \tan \theta}\right\}^2 \\ & =\left\{\frac{n \tan \theta\left(\sec ^n \theta+\cos ^n \theta\right.}{\tan \theta(\sec \theta+\cos \theta)}\right\}^2 \\ & =\frac{\mathrm{n}^2\left(\sec ^{\mathrm{n}} \theta+\cos ^{\mathrm{n}} \theta\right)^2}{(\sec \theta+\cos \theta)^2} \\ & \end{aligned}\)
\(=\frac{\mathrm{n}^2\left[\mathrm{y}^2+4\right]}{\mathrm{x}^2+4}\)