Consider the following for the two (02) items that follow: Let x =...

Given $x = \sec\theta - \cos\theta$ and $y = \sec^4\theta - \cos^4\theta$.

Differentiating $x$ with respect to $\theta$:
$\dfrac{dx}{d\theta} = \sec\theta \tan\theta + \sin\theta$
$\dfrac{dx}{d\theta} = \tan\theta (\sec\theta + \cos\theta)$

Differentiating $y$ with respect to $\theta$:
$\dfrac{dy}{d\theta} = 4\sec^3\theta (\sec\theta \tan\theta) - 4\cos^3\theta (-\sin\theta)$
$\dfrac{dy}{d\theta} = 4\sec^4\theta \tan\theta + 4\cos^3\theta \sin\theta$
$\dfrac{dy}{d\theta} = 4\tan\theta (\sec^4\theta + \cos^4\theta)$

Now, $\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}} = \dfrac{4\tan\theta (\sec^4\theta + \cos^4\theta)}{\tan\theta (\sec\theta + \cos\theta)} = \dfrac{4(\sec^4\theta + \cos^4\theta)}{\sec\theta + \cos\theta}$

Squaring both sides:
\left(\dfrac{dy}{dx}\right)^2 = \dfrac{16(\sec^4\theta + \cos^4\theta)^2}{(\sec\theta + \cos\theta)^2}$<br><br>Using the algebraic identity$(a+b)^2 = (a-b)^2 + 4ab$, we can rewrite the numerator and denominator:<br>$(\sec\theta + \cos\theta)^2 = (\sec\theta - \cos\theta)^2 + 4\sec\theta\cos\theta = x^2 + 4$<br>$(\sec^4\theta + \cos^4\theta)^2 = (\sec^4\theta - \cos^4\theta)^2 + 4\sec^4\theta\cos^4\theta = y^2 + 4$<br><br>Substituting these values, we get:<br>\left(\dfrac{dy}{dx}\right)^2 = \dfrac{16(y^2 + 4)}{x^2 + 4}$

Answer: $\dfrac{16(y^2+4)}{(x^2+4)}$