If x=a≤ft[≤ft\{ t+1/2 ≤ft( ² t/2)\}] and y=a t then d y/d x=

Correct Answer is : tan⁡t ttant

Correct Answer is : tan⁡t ttant

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COMEDK 2025 Evening Solved Paper

Question : 160 of 180

Marks: +1, -0

x=a\left[\left\{\cos t+\frac{1}{2} \log \left(\tan ^2 \frac{t}{2}\right)\right\}\right]

y=a \sin t

\frac{d y}{d x}=

Solution:

Concept:

Parametric differentiation using individual time derivatives combined via the chain rule.

Explanation:

Given

x = a\Bigl(\cos t + \frac12\log(\tan^2\frac t2)\Bigr)

and

y = a\sin t

Simplify

x

using

\frac12\log(\tan^2\frac t2) = \log(\tan\frac t2)

for positive arguments.

Differentiate both with respect to

t

\frac{dy}{dt} = a\cos t

For

x

\frac{dx}{dt} = a\Bigl(-\sin t + \frac{d}{dt}\log(\tan\frac t2)\Bigr)

Compute

\frac{d}{dt}\log(\tan\frac t2) = \frac{1}{\tan(t/2)}\cdot\sec^2(t/2)\cdot\frac12 = \frac{1}{2\sin(t/2)\cos(t/2)} = \frac{1}{\sin t}

Thus

\frac{dx}{dt} = a\Bigl(-\sin t + \frac{1}{\sin t}\Bigr) = a\frac{1-\sin^2t}{\sin t} = a\frac{\cos^2t}{\sin t}

Now

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a\cos t}{a\cos^2t/\sin t} = \frac{\cos t\sin t}{\cos^2t} = \tan t

Answer:

Option B:

\tan t

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