If x = a sin theta and y = b cos theta,then f | vedclass

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(C) Given $x = a \sin 	heta$ and $y = b \cos 	heta$. First,find the derivatives with respect to $	heta$: $\frac{dx}{d	heta} = a \cos 	heta$ and $

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$-\frac{b}{a^2} \sec^3 	heta$

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(C) Given $x = a \sin 	heta$ and $y = b \cos 	heta$. First,find the derivatives with respect to $	heta$: $\frac{dx}{d	heta} = a \cos 	heta$ and $\frac{dy}{d	heta} = -b \sin 	heta$. Now,find $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{-b \sin 	heta}{a \cos 	heta} = -\frac{b}{a} 	an 	heta$. Next,differentiate $\frac{dy}{dx}$ with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{dx} \left( -\frac{b}{a} 	an 	heta ight) = -\frac{b}{a} \sec^2 	heta \cdot \frac{d	heta}{dx}$. Since $\frac{dx}{d	heta} = a \cos 	heta$,we have $\frac{d	heta}{dx} = \frac{1}{a \cos 	heta}$. Substituting this value: $\frac{d^2y}{dx^2} = -\frac{b}{a} \sec^2 	heta \cdot \frac{1}{a \cos 	heta} = -\frac{b}{a^2} \sec^3 	heta$.

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 $x = a \sin \theta$ and $y = b \cos \theta$,then $\frac{d^2y}{dx^2}$ is

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