Question:

Given $x = \sec\theta - \cos\theta$ and $y = \sec^n\theta - \cos^n\theta$. We aim to find $k$ such that $\left(x^2+4\right)\!\left(\dfrac{dy}{dx}\right)^{\!2} = n^2(y^2+k)$. First, we find the derivatives of $x$ and $y$ with respect to $\theta$: $$ \frac{dx}{d\theta} = \frac{d}{d\theta}(\sec\theta - \cos\theta) = \sec\theta\tan\theta + \sin\theta $$ This expression can be factored as: $$ \frac{dx}{d\theta} = \frac{\sin\theta}{\cos^2\theta} + \sin\theta = \sin\theta\left(\frac{1}{\cos^2\theta} + 1\right) = \sin\theta(\sec^2\theta+1) $$ Alternatively, by factoring $\tan\theta$: $$ \frac{dx}{d\theta} = \tan\theta\left(\sec\theta + \frac{\sin\theta}{\tan\theta}\right) = \tan\theta\left(\sec\theta + \frac{\sin\theta}{\sin\theta/\cos\theta}\right) = \tan\theta(\sec\theta+\cos\theta) $$ Next, we find $\frac{dy}{d\theta}$: $$ \frac{dy}{d\theta} = \frac{d}{d\theta}(\sec^n\theta - \cos^n\theta) = n\sec^{n-1}\theta(\sec\theta\tan\theta) - n\cos^{n-1}\theta(-\sin\theta) $$ $$ \frac{dy}{d\theta} = n\sec^n\theta\tan\theta + n\cos^{n-1}\theta\sin\theta $$ This expression can be factored as: $$ \frac{dy}{d\theta} = n\tan\theta\left(\sec^n\theta + \frac{\cos^{n-1}\theta\sin\theta}{\tan\theta}\right) = n\tan\theta\left(\sec^n\theta + \frac{\cos^{n-1}\theta\sin\theta}{\sin\theta/\cos\theta}\right) $$ $$ \frac{dy}{d\theta} = n\tan\theta(\sec^n\theta + \cos^n\theta) $$ Now, we determine $\frac{dy}{dx}$ using the chain rule: $$ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{n\tan\theta(\sec^n\theta+\cos^n\theta)}{\tan\theta(\sec\theta+\cos\theta)} = \frac{n(\sec^n\theta+\cos^n\theta)}{\sec\theta+\cos\theta} $$ Next, we simplify the terms $x^2+4$ and $y^2+4$. For $x^2+4$: $$ x^2+4 = (\sec\theta-\cos\theta)^2+4 = (\sec^2\theta - 2\sec\theta\cos\theta + \cos^2\theta) + 4 $$ Since $\sec\theta\cos\theta = 1$, this simplifies to: $$ x^2+4 = \sec^2\theta - 2 + \cos^2\theta + 4 = \sec^2\theta + 2 + \cos^2\theta $$ This expression is equivalent to $(\sec\theta+\cos\theta)^2$: $$ (\sec\theta+\cos\theta)^2 = \sec^2\theta + 2\sec\theta\cos\theta + \cos^2\theta = \sec^2\theta + 2 + \cos^2\theta $$ Thus, $x^2+4 = (\sec\theta+\cos\theta)^2$. Similarly, for $y^2+4$: $$ y^2+4 = (\sec^n\theta-\cos^n\theta)^2+4 = (\sec^{2n}\theta - 2\sec^n\theta\cos^n\theta + \cos^{2n}\theta) + 4 $$ Since $\sec^n\theta\cos^n\theta = (\sec\theta\cos\theta)^n = 1^n = 1$, this simplifies to: $$ y^2+4 = \sec^{2n}\theta - 2 + \cos^{2n}\theta + 4 = \sec^{2n}\theta + 2 + \cos^{2n}\theta $$ This expression is equivalent to $(\sec^n\theta+\cos^n\theta)^2$: $$ (\sec^n\theta+\cos^n\theta)^2 = \sec^{2n}\theta + 2\sec^n\theta\cos^n\theta + \cos^{2n}\theta = \sec^{2n}\theta + 2 + \cos^{2n}\theta $$ Thus, $y^2+4 = (\sec^n\theta+\cos^n\theta)^2$. Now, we substitute these simplified expressions into the given equation $\left(x^2+4\right)\!\left(\dfrac{dy}{dx}\right)^{\!2}$: $$ \left(x^2+4\right)\!\left(\dfrac{dy}{dx}\right)^{\!2} = (\sec\theta+\cos\theta)^2 \left(\frac{n(\sec^n\theta+\cos^n\theta)}{\sec\theta+\cos\theta}\right)^2 $$ $$ \left(x^2+4\right)\!\left(\dfrac{dy}{dx}\right)^{\!2} = (\sec\theta+\cos\theta)^2 \frac{n^2(\sec^n\theta+\cos^n\theta)^2}{(\sec\theta+\cos\theta)^2} $$ The term $(\sec\theta+\cos\theta)^2$ cancels out: $$ \left(x^2+4\right)\!\left(\dfrac{dy}{dx}\right)^{\!2} = n^2(\sec^n\theta+\cos^n\theta)^2 $$ Using the identity $y^2+4 = (\sec^n\theta+\cos^n\theta)^2$, we substitute this back into the equation: $$ \left(x^2+4\right)\!\left(\dfrac{dy}{dx}\right)^{\!2} = n^2(y^2+4) $$ Comparing this result with the given form $n^2(y^2+k)$, we conclude that $k=4$.