CUET UG 2025 — Mathematics Calculus

Given $x = a\sec^3\theta$x=asec3θ and $y = a\tan^3\theta$y=atan3θ.

Since both $x$x and $y$y are in terms of $\theta$θ, use parametric differentiation:

$$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$$dxdy​=dx/dθdy/dθ​

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For $x = a\sec^3\theta$x=asec3θ:

$$\frac{dx}{d\theta} = 3a\sec^2\theta \cdot \sec\theta\tan\theta$$dθdx​=3asec2θ⋅secθtanθ

$$\frac{dx}{d\theta} = 3a\sec^3\theta\tan\theta$$dθdx​=3asec3θtanθ

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For $y = a\tan^3\theta$y=atan3θ:

$$\frac{dy}{d\theta} = 3a\tan^2\theta \cdot \sec^2\theta$$dθdy​=3atan2θ⋅sec2θ

$$\frac{dy}{d\theta} = 3a\tan^2\theta\sec^2\theta$$dθdy​=3atan2θsec2θ

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$$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$$dxdy​=dx/dθdy/dθ​

$$\frac{dy}{dx} = \frac{3a\tan^2\theta\sec^2\theta}{3a\sec^3\theta\tan\theta}$$dxdy​=3asec3θtanθ3atan2θsec2θ​

$$\frac{dy}{dx} = \frac{\tan^2\theta\sec^2\theta}{\sec^3\theta\tan\theta}$$dxdy​=sec3θtanθtan2θsec2θ​

$$\frac{dy}{dx} = \frac{\tan\theta}{\sec\theta}$$dxdy​=secθtanθ​

$$\frac{dy}{dx} = \frac{\sin\theta/\cos\theta}{1/\cos\theta}$$dxdy​=1/cosθsinθ/cosθ​

$$\frac{dy}{dx} = \sin\theta$$dxdy​=sinθ

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At $\theta = \frac{\pi}{3}$θ=3π​:

$$\frac{dy}{dx}\bigg|_{\theta=\pi/3} = \sin\left(\frac{\pi}{3}\right)$$dxdy​​θ=π/3​=sin(3π​)

$$\frac{dy}{dx}\bigg|_{\theta=\pi/3} = \frac{\sqrt{3}}{2}$$dxdy​​θ=π/3​=23​​