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(D) Given $x = \cos 	heta + \log 	an \frac{	heta}{2}$.On differentiating with respect to $	heta$,we get:$\frac{dx}{d	heta} = -\sin 	heta + \frac

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$	heta = n \pi, n \in Z$

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(D) Given $x = \cos 	heta + \log 	an \frac{	heta}{2}$.On differentiating with respect to $	heta$,we get:$\frac{dx}{d	heta} = -\sin 	heta + \frac{1}{	an \frac{	heta}{2}} \cdot \sec^2 \frac{	heta}{2} \cdot \frac{1}{2}$$= -\sin 	heta + \frac{\cos \frac{	heta}{2}}{\sin \frac{	heta}{2}} \cdot \frac{1}{\cos^2 \frac{	heta}{2}} \cdot \frac{1}{2}$$= -\sin 	heta + \frac{1}{2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}} = -\sin 	heta + \frac{1}{\sin 	heta}$$= \frac{1 - \sin^2 	heta}{\sin 	heta} = \frac{\cos^2 	heta}{\sin 	heta} \quad ...(i)$Now,$y = \sin 	heta$. On differentiating with respect to $	heta$,we get:$\frac{dy}{d	heta} = \cos 	heta \quad ...(ii)$Therefore,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{\cos 	heta}{\frac{\cos^2 	heta}{\sin 	heta}} = 	an 	heta$.For $\frac{dy}{dx} = 0$,we must have $	an 	heta = 0$.This implies $	heta = n \pi$ for $n \in Z$.

If the parametric equations of a curve are given by $x = \cos \theta + \log \tan \frac{\theta}{2}$ and $y = \sin \theta$,then the points for which $\frac{dy}{dx} = 0$ are given by

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