If the parametric equation of curve is given by \(x=\cos \theta+\log \tan \frac{\theta}{2}\) and \(y=\sin \theta\), then the points for which \(\frac{d y}{d x}=0\) are given by

\(x=\cos \theta+\log \tan \frac{\theta}{2}\)
Oñ đifferrēñtiāting ̄ w.r.t. \(\theta\), wē gẹt
\(\frac{d x}{d \theta}=-\sin \theta+\frac{1}{\tan \frac{\theta}{2}} \times \sec ^{2} \frac{\theta}{2} \times \frac{1}{2}\)
\(\begin{aligned}
&=-\sin \theta+\frac{1}{\sin \theta} \\
&=\frac{1-\sin ^{2} \theta}{\sin \theta} \\
&=\frac{\cos ^{2} \theta}{\sin \theta} ...(i)
\end{aligned}\)
Now, \(y=\sin \theta\)
On differentiating w.r.t. \(\theta\),
\(\frac{d y}{d \theta}=\cos \theta...(ii)\)
\(\begin{aligned}
\therefore \quad \frac{d y}{d x} &=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}} \\
&=\frac{\cos \theta}{\frac{\cos ^{2} \theta}{\sin \theta}}=\tan \theta
\end{aligned}\)
[from Eqs. (i) and (ii)]
If \(\frac{d y}{d x}=0\), then \(\tan \theta=0\).
Hence, \(\theta=n \pi, n \in Z\).