CUET UG 2025 — Mathematics Calculus

Since both $x$x and $y$y are given in terms of a parameter $t$t:

$$\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$$dxdy​=dtdx​dtdy​​

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$$y = a\sin t$$y=asint

$$\dfrac{dy}{dt} = a\cos t$$dtdy​=acost

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$$x = a\left(\cos t + \log\tan\dfrac{t}{2}\right)$$x=a(cost+logtan2t​)

$$\dfrac{dx}{dt} = a\left(-\sin t + \dfrac{1}{\tan\frac{t}{2}} \cdot \sec^2\dfrac{t}{2} \cdot \dfrac{1}{2}\right)$$dtdx​=a(−sint+tan2t​1​⋅sec22t​⋅21​)

The second term inside the bracket simplifies as:

$$\dfrac{\cos\frac{t}{2}}{\sin\frac{t}{2}} \cdot \dfrac{1}{\cos^2\frac{t}{2}} \cdot \dfrac{1}{2}$$sin2t​cos2t​​⋅cos22t​1​⋅21​

$$= \dfrac{1}{2\sin\frac{t}{2}\cos\frac{t}{2}}$$=2sin2t​cos2t​1​

$$= \dfrac{1}{\sin t}$$=sint1​

So:

$$\dfrac{dx}{dt} = a\left(-\sin t + \dfrac{1}{\sin t}\right)$$dtdx​=a(−sint+sint1​)

$$= a\left(\dfrac{-\sin^2 t + 1}{\sin t}\right)$$=a(sint−sin2t+1​)

$$= a\left(\dfrac{\cos^2 t}{\sin t}\right)$$=a(sintcos2t​)

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$$\dfrac{dy}{dx} = \dfrac{a\cos t}{\;a\cdot\dfrac{\cos^2 t}{\sin t}\;}$$dxdy​=a⋅sintcos2t​acost​

$$= \dfrac{\cos t \cdot \sin t}{\cos^2 t}$$=cos2tcost⋅sint​

$$= \dfrac{\sin t}{\cos t}$$=costsint​

$$= \tan t$$=tant

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At $t = \dfrac{\pi}{4}$t=4π​:

$$\dfrac{dy}{dx}\Bigg|_{t=\pi/4} = \tan\dfrac{\pi}{4} = \boxed{1}$$dxdy​​t=π/4​=tan4π​=1​