Let \(x = \sec \theta - \cos \theta \text{ and } y = \sec^4 \theta - \cos^4 \theta \\ \) What is \((\frac{dy}{dx})^2\) equal to?

Calculation:

Given,

\(x = \sec\theta - \cos\theta\)

\(y = \sec^4\theta - \cos^4\theta\)

Compute derivatives w.r.t. \(\theta\):

\(\dfrac{dx}{d\theta} = \sec\theta\,\tan\theta + \sin\theta\)

\(\dfrac{dy}{d\theta} = 4\sec^4\theta\,\tan\theta \;+\; 4\cos^3\theta\,\sin\theta\)

From the ratio \(\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\)

\(\dfrac{dy}{dx} = \frac{4\bigl(\sec^4\theta\,\tan\theta + \cos^3\theta\,\sin\theta\bigr)} {\sec\theta\,\tan\theta + \sin\theta}\)

Using identities \(x^2 = \sec^2\theta + \cos^2\theta - 2\), \(\sec^3\theta + \cos^5\theta)^2 = y^2 + 4\), and \((1 + \cos^2\theta)^2 = x^2 + 4,\)

\(\displaystyle \Bigl(\frac{dy}{dx}\Bigr)^{2} = 16 \,\frac{\,y^{2} + 4\,}{\,x^{2} + 4\,} \)

∴  \(\displaystyle \Bigl(\frac{dy}{dx}\Bigr)^{2} = 16\,\frac{y^{2} + 4}{x^{2} + 4} \)

Hence, the correct answer is Option 3.