Step 1: Set up shorter symbols.
Let $s=\sec\theta$ and $c=\cos\theta$, so $sc=1$. Then $x=s+c$ and $y=s^n+c^n$.
Step 2: Square both.
$x^2 = s^2+2+c^2$, so $x^2-4 = s^2-2+c^2 = (s-c)^2$. In the same way $y^2-4 = (s^n-c^n)^2$.
Step 3: Use the derivatives of s and c.
$\frac{ds}{d\theta} = s\tan\theta$ and $\frac{dc}{d\theta} = -\sin\theta = -c\tan\theta$.
Step 4: Find dx and dy.
\[ \frac{dx}{d\theta} = \tan\theta\,(s-c), \qquad \frac{dy}{d\theta} = n\tan\theta\,(s^n-c^n) \]
Step 5: Divide.
\[ \frac{dy}{dx} = n\cdot\frac{s^n-c^n}{s-c} = n\sqrt{\frac{y^2-4}{x^2-4}} \]
The ratio of the two brackets is taken as positive, matching the option.
Final Answer:
So $\frac{dy}{dx}= n\sqrt{\frac{y^2-4}{x^2-4}}$.
\[ \boxed{n\sqrt{\dfrac{y^2-4}{x^2-4}}} \]