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

$$\begin{gathered} \mathrm{x}=\mathrm{sec\theta }-\mathrm{cos\theta }, \mathrm{y}={\sec }^{10}\mathrm{\theta }-{\cos }^{10}\mathrm{\theta } \\ \frac{\mathrm{dx}}{\mathrm{d\theta }}=\mathrm{sec\theta } \mathrm{tan\theta }+\mathrm{sin\theta }=\mathrm{tan\theta }\left(\mathrm{sec\theta }+\mathrm{cos\theta }\right) \\ \frac{\mathrm{dy}}{\mathrm{d\theta }}=10{\sec }^{10}\mathrm{\theta } \mathrm{tan\theta }+10{\cos }^9\mathrm{\theta }.\mathrm{sin\theta } \\ =10 \mathrm{tan\theta } \left({\sec }^{10}\mathrm{\theta }+{\cos }^{10}\mathrm{\theta }\right) \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{10\tan \mathrm{\theta }({\sec }^{10}\mathrm{\theta }+{\cos }^{10}\mathrm{\theta })}{\mathrm{tan\theta }\left(\mathrm{sec\theta }+\mathrm{cos\theta }\right)} \end{gathered}$$

$$\begin{gathered} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{10\sqrt{{\mathrm{y}}^2+4}}{\sqrt{{\mathrm{x}}^2+4}} \\ \left({\mathrm{x}}^2+4\right){\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2=100\left({\mathrm{y}}^2+4\right) \\ \mathrm{k}=100 \end{gathered}$$