If x=a(t-sin t) and y=a(1+cos t), then frac{d | vedclass

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(A) Given, $x=a(t-\sin t)$ and $y=a(1+\cos t)$.First, differentiate $x$ and $y$ with respect to $t$:$\frac{dx}{dt} = a(1-\cos t)$ and $\frac{dy}{dt} =

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$\frac{1}{4 a \sin ^4(\frac{t}{2})}$

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(A) Given, $x=a(t-\sin t)$ and $y=a(1+\cos t)$.First, differentiate $x$ and $y$ with respect to $t$:$\frac{dx}{dt} = a(1-\cos t)$ and $\frac{dy}{dt} = -a \sin t$.Now, find $\frac{dy}{dx}$:$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-a \sin t}{a(1-\cos t)} = \frac{-2 \sin(t/2) \cos(t/2)}{2 \sin^2(t/2)} = -\cot(t/2)$.Next, find the second derivative $\frac{d^2y}{dx^2}$:$\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx} = \frac{d}{dt}(-\cot(t/2)) \cdot \frac{1}{a(1-\cos t)}$.$\frac{d^2y}{dx^2} = -(-\csc^2(t/2) \cdot \frac{1}{2}) \cdot \frac{1}{a(2 \sin^2(t/2))} = \frac{\csc^2(t/2)}{4a \sin^2(t/2)}$.Since $\csc^2(t/2) = \frac{1}{\sin^2(t/2)}$, we get:$\frac{d^2y}{dx^2} = \frac{1}{4a \sin^4(t/2)}$.Thus, option $A$ is correct.

List-$I$	List-$II$
$(i)$ $x = a(\theta - \sin \theta), y = a(1 - \cos \theta)$	$(A)$ $4\sqrt{3}$
(ii) $x = 3\cos \theta - 2\cos^3 \theta, y = 3\sin \theta - 2\sin^3 \theta$	$(B)$ $-\frac{1}{3\sqrt{3}}$
(iii) $x = 3\cos \theta - \cos^3 \theta, y = 3\sin \theta - \sin^3 \theta$	$(C)$ $\sqrt{3}$
(iv) $x = a \log \sin \theta, y = a \tan \theta$	$(D)$ $\frac{1}{\sqrt{3}}$
	$(E)$ $\frac{1}{3\sqrt{3}}$

If $x=a(t-\sin t)$ and $y=a(1+\cos t)$, then $\frac{d^2 y}{d x^2}=$

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