If \(x=\sec \theta-\cos \theta, y=\sec ^{10} \theta-\cos ^{10} \theta\) and \(\left(x^2+4\right)\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2=\mathrm{k}\left(y^2+4\right)\), then the value of k is

\(\begin{aligned} & x=\sec \theta-\cos \theta \text { and } y=\sec ^{10} \theta-\cos ^{10} \theta \\ \therefore \quad & \frac{\mathrm{~d} x}{\mathrm{~d} \theta}=\sec \theta \tan \theta+\sin \theta \text { and } \\ & \frac{\mathrm{d} y}{\mathrm{~d} \theta}=10 \sec ^9 \theta \cdot \sec \theta \tan \theta-10 \cos ^9 \theta \cdot(-\sin \theta)\end{aligned}\)
\(\begin{aligned}
& =10 \sec ^{10} \theta \tan \theta+10 \cos ^9 \theta \sin \theta \\
& \therefore \quad \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{~d} \theta}}{\frac{\mathrm{~d} x}{\mathrm{~d} \theta}}=\frac{10 \sec ^{10} \theta \tan \theta+10 \cos ^9 \theta \sin \theta}{\sec \theta \tan \theta+\sin \theta}
\end{aligned}\)
\(=\frac{10\left(\sec ^{10} \theta+\cos ^{10} \theta\right)}{\sec \theta+\cos \theta}\)
\(\begin{aligned} \therefore \quad\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2 & =\frac{10^2\left(\sec ^{10} \theta+\cos ^{10} \theta\right)^2}{(\sec \theta+\cos \theta)^2} \\ & =\frac{100\left[\left(\sec ^{10} \theta-\cos ^{10} \theta\right)^2+4 \sec ^{10} \theta \cos ^{10} \theta\right]}{(\sec \theta-\cos \theta)^2+4 \sec \theta \cos \theta}\end{aligned}\)
\(\begin{aligned} \therefore \quad & \left(\frac{d y}{d x}\right)^2=\frac{100\left(y^2+4\right)}{x^2+4} \\ & \Rightarrow\left(x^2+4\right)\left(\frac{d y}{d x}\right)^2=100\left(y^2+4\right) \Rightarrow \mathrm{k}=100\end{aligned}\)