$\mathrm{If} \mathrm{x}=\sec \mathrm{\theta }-\cos \mathrm{\theta }, \mathrm{y}={\sec }^{10}\mathrm{\theta }-{\cos }^{10}\mathrm{\theta }\text{ and }\left({\mathrm{x}}^2+4\right){\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2=\mathrm{k}\left({\mathrm{y}}^2+4\right),\text{ then }\mathrm{k}=$

$$\begin{gathered} \begin{array}{l}\mathrm{x}=\sec \mathrm{\theta }-\cos \mathrm{\theta }\Rightarrow \frac{\mathrm{dx}}{\mathrm{d\theta }}=\sec \mathrm{\theta tan}\mathrm{\theta }+\sin \mathrm{\theta } \\ =\tan \mathrm{\theta }(\sec \mathrm{\theta }+\cos \mathrm{\theta })=\tan \mathrm{\theta }\sqrt{{\mathrm{x}}^2+4}\\ \mathrm{y}={\sec }^{10}\mathrm{\theta }-{\cos }^{10}\mathrm{\theta } \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{d\theta }}=10{\sec }^9\mathrm{\theta sec}\mathrm{\theta tan}\mathrm{\theta }+10{\cos }^9\mathrm{\theta sin}\mathrm{\theta }\\ =10\tan \mathrm{\theta }\left({\sec }^{10}\mathrm{\theta }+{\cos }^{10}\mathrm{\theta }\right)=10\tan \mathrm{\theta }\sqrt{{\mathrm{y}}^2+4}.\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{d\theta }}/\frac{\mathrm{dx}}{\mathrm{d\theta }}=\frac{10\tan \mathrm{\theta }\sqrt{{\mathrm{y}}^2+4}}{\tan \mathrm{\theta }\sqrt{{\mathrm{x}}^2+4}} \\ =10\frac{\sqrt{{\mathrm{y}}^2+4}}{\sqrt{{\mathrm{x}}^2+4}}\Rightarrow \left({\mathrm{x}}^2+4\right){\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2=100\left({\mathrm{y}}^2+4\right)\Rightarrow \mathrm{k}=100.\end{array} \end{gathered}$$