If x=e^{theta} sin theta and y=e^{theta} cos | vedclass

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(A) Given,$x=e^{	heta} \sin 	heta$ and $y=e^{	heta} \cos 	heta$. We differentiate both with respect to $	heta$: $\frac{dx}{d	heta} = e^{	heta}

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(A) Given,$x=e^{	heta} \sin 	heta$ and $y=e^{	heta} \cos 	heta$. We differentiate both with respect to $	heta$: $\frac{dx}{d	heta} = e^{	heta} \sin 	heta + e^{	heta} \cos 	heta = e^{	heta}(\sin 	heta + \cos 	heta)$ $\frac{dy}{d	heta} = e^{	heta} \cos 	heta - e^{	heta} \sin 	heta = e^{	heta}(\cos 	heta - \sin 	heta)$ Now,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{e^{	heta}(\cos 	heta - \sin 	heta)}{e^{	heta}(\cos 	heta + \sin 	heta)} = \frac{\cos 	heta - \sin 	heta}{\cos 	heta + \sin 	heta}$. Dividing numerator and denominator by $\cos 	heta$,we get $\frac{dy}{dx} = \frac{1 - 	an 	heta}{1 + 	an 	heta}$. At $(x, y) = (1, 1)$,we have $x = e^{	heta} \sin 	heta = 1$ and $y = e^{	heta} \cos 	heta = 1$. Dividing these,$\frac{y}{x} = \frac{e^{	heta} \cos 	heta}{e^{	heta} \sin 	heta} = \cot 	heta = 1$,so $	an 	heta = 1$. Substituting $	an 	heta = 1$ into the expression for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0$.

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 $x=e^{\theta} \sin \theta$ and $y=e^{\theta} \cos \theta$,where $\theta$ is a parameter,then $\frac{dy}{dx}$ at $(1,1)$ is equal to

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