If $x = 3 \cos \theta - 2 \cos^3 \theta$ and $y = 3 \sin \theta - 2 \sin^3 \theta$, then $\frac{dy}{dx}$ is equal to:

$\cot \theta$

The correct answer is Option (3) → $\cot \theta$ ##

We have, $x = 3 \cos \theta - 2 \cos^3 \theta$ and $y = 3 \sin \theta - 2 \sin^3 \theta$

Taking derivative of both equations w.r.t. $\theta$, we get

$∴\frac{dx}{d\theta} = \frac{d}{d\theta} (3 \cos \theta) - \frac{d}{d\theta} (2 \cos^3 \theta)$

$= 3 \cdot (-\sin \theta) - 2 \cdot 3 \cos^2 \theta \cdot \frac{d}{d\theta} \cos \theta$

$= -3 \sin \theta + 6 \cos^2 \theta \sin \theta \quad \left[ ∵\frac{d}{dx} \cos x = -\sin x \right] \dots(i)$

and $\frac{dy}{d\theta} = \frac{d}{d\theta} (3 \sin \theta) - \frac{d}{d\theta} (2 \sin^3 \theta)$

$\Rightarrow \frac{dy}{d\theta} = 3 \cos \theta - 2 \cdot 3 \sin^2 \theta \cdot \frac{d}{d\theta} \sin \theta$

$= 3 \cos \theta - 6 \sin^2 \theta \cos \theta \quad \dots(ii)$

On dividing Eqs. (i) and (ii), we get:

$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3 \cos \theta - 6 \sin^2 \theta \cos \theta}{-3 \sin \theta + 6 \cos^2 \theta \sin \theta}$

$= \frac{3 \cos \theta (1 - 2\sin^2 \theta)}{3 \sin \theta (-1 + 2\cos^2 \theta)} = \cot \theta \cdot \frac{\cos 2\theta}{\cos 2\theta} = \cot \theta$ $[∵\cos 2\theta = 1 - 2\sin^2 \theta = 2\cos^2 \theta - 1]$