If $x = a \sec^3θ, y = a \tan^3θ$, then $\frac{d^2y}{dx^2}$ equals.

$\frac{\cos^5θ}{3a\sin θ}$

The correct answer is Option (1) → $\frac{\cos^5θ}{3a\sin θ}$

Given: $x = a \sec^3\theta$, $y = a \tan^3\theta$

Differentiate $x$ with respect to $\theta$:

$\frac{dx}{d\theta} = a \cdot 3\sec^2\theta \cdot \sec\theta \tan\theta = 3a \sec^3\theta \tan\theta$

$\frac{dy}{d\theta} = a \cdot 3\tan^2\theta \cdot \sec^2\theta = 3a \tan^2\theta \sec^2\theta$

Now, first derivative:

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{3a \tan^2\theta \sec^2\theta}{3a \sec^3\theta \tan\theta} = \frac{\tan\theta \sec^2\theta}{\sec^3\theta} = \frac{\tan\theta}{\sec\theta} = \sin\theta$

Now, second derivative:

$\frac{d^2y}{dx^2} = \frac{d}{d\theta}(\sin\theta) \cdot \frac{d\theta}{dx}$

$= \cos\theta \cdot \frac{1}{\frac{dx}{d\theta}} = \cos\theta \cdot \frac{1}{3a \sec^3\theta \tan\theta}$

$= \frac{\cos\theta}{3a \cdot \frac{1}{\cos^3\theta} \cdot \frac{\sin\theta}{\cos\theta}} = \frac{\cos\theta}{3a \cdot \frac{\sin\theta}{\cos^4\theta}} = \frac{\cos^5\theta}{3a \sin\theta}$