(b) We have,
$$
x=\sec \theta-\cos \theta
$$
differential w.r.t $\theta$
$$
\begin{aligned}
\frac{d x}{d \theta} & =\sec \theta \tan \theta+\sin \theta \\
& =\sec \theta \cdot \frac{\sin \theta}{\cos \theta}+\sin \theta \\
& =\tan \theta(\sec \theta+\cos \theta)
\end{aligned}
$$
Now, $\quad y=\sec ^n \theta-\cos ^n \theta$
differential w.r.t ' $\theta$ '
$$
\begin{aligned}
\frac{d y}{d \theta} & =n \sec ^{n-1} \theta \cdot \sec \theta \tan \theta-n \cos ^{n-1} \theta(-\sin \theta) \\
& =n \sec ^n \theta \frac{\sin \theta}{\cos \theta}+n \cos ^{n-1} \theta \cdot \sin \theta \\
& =n \frac{\sin \theta}{\cos \theta}\left(\sec ^n \theta+\cos ^n \theta\right) \\
& =n \tan \theta\left(\sec ^n \theta+\cos ^n \theta\right)
\end{aligned}
$$
Now,
$$
\begin{aligned}
& \frac{d y}{d x}=\frac{n \tan \theta\left(\sec ^n \theta+\cos ^n \theta\right)}{\tan \theta(\sec \theta+\cos \theta)} \\
& \frac{d y}{d x}=\frac{n\left(\sec ^n \theta+\cos ^n \theta\right)}{(\sec \theta+\cos \theta)}
\end{aligned}
$$
Squaring both side
$\begin{aligned} & \left(\frac{d y}{d x}\right)^2=\frac{n^2\left(\sec ^n \theta+\cos ^n \theta\right)^2}{(\sec \theta+\cos \theta)^2} \\ & \left(\frac{d y}{d x}\right)^2=\frac{n^2\left(\sec ^{2 n} \theta+\cos ^{2 n} \theta+2\right)}{\left(\sec ^2 \theta+\cos ^2 \theta+2\right)} \\ & =\frac{n^2\left(\sec ^{2 n} \theta+\cos ^{2 n} \theta-2+4\right)}{\left(\sec ^2 \theta+\cos ^2 \theta-2+4\right)} \\ & =\frac{n^2\left[\left(\sec ^n \theta-\cos ^n \theta\right)^2+4\right]}{(\sec \theta-\cos \theta)^2+4} \\ & \left(\frac{d y}{d x}\right)^2=\frac{n^2\left(y^2+4\right)}{\left(x^2+4\right)} \\ & \Rightarrow \frac{d y}{d x}=n \sqrt{\frac{y^2+4}{x^2+4}} \\ & \end{aligned}$