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

We have,
$$\begin{gathered} \frac{\mathrm{dx}}{\mathrm{d\theta }}=-\sin \mathrm{\theta }+\frac{{\sec }^2\mathrm{\theta }/2}{2\tan \mathrm{\theta }/2}=-\sin \mathrm{\theta }+\frac{1}{2\sin \frac{\mathrm{\theta }}{2}\cos \frac{\mathrm{\theta }}{2}} \\ =-\sin \mathrm{\theta }+\frac{1}{\sin \mathrm{\theta }}=\frac{1-{\sin }^2\mathrm{\theta }}{\sin \mathrm{\theta }}=\frac{{\cos }^2\mathrm{\theta }}{\sin \mathrm{\theta }} \end{gathered}$$
and $\frac{\mathrm{dy}}{\mathrm{d\theta }}=\cos \mathrm{\theta }$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}/\mathrm{d\theta }}{\mathrm{dx}/\mathrm{d\theta }}=\frac{\cos \mathrm{\theta sin}\mathrm{\theta }}{{\cos }^2\mathrm{\theta }}=\tan \mathrm{\theta }$
Also,  $\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}={\sec }^2\mathrm{\theta }\cdot \frac{\mathrm{d\theta }}{\mathrm{dx}}=\frac{\sin \mathrm{\theta }}{{\cos }^4\mathrm{\theta }}$
$\therefore \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=0\Rightarrow \sin \mathrm{\theta }=0\Rightarrow \mathrm{\theta }=\mathrm{n\pi },\mathrm{n}\in \mathrm{Z}$