If x = asec^3θ, y = a tan^3θ, then dy/dx at θ=(π)/3 is

The correct answer is Option (1) → $\frac{\sqrt{3}}{2}$

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

Compute derivatives w.r.t $\theta$:

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

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

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

At $\theta = \pi/3$: $\sin (\pi/3) = \frac{\sqrt{3}}{2}$

Answer: $\frac{\sqrt{3}}{2}$