The equation of the normal to the curve \(x=\theta+\sin \theta, y=1+\cos \theta\) at \(\theta=\frac{\pi}{2}\) is

\(\begin{aligned}
& x=\theta+\sin \theta, y=1+\cos \theta \\
& \frac{\mathrm{d} x}{\mathrm{~d} \theta}=1+\cos \theta, \frac{\mathrm{d} y}{\mathrm{~d} \theta}=-\sin \theta \\
& \therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{~d} \theta}}{\frac{\mathrm{~d} x}{\mathrm{~d} \theta}}=\frac{-\sin \theta}{1+\cos \theta} \\
& \left.\therefore \quad \frac{\mathrm{d} y}{\mathrm{dx}}\right|_{\theta=\frac{\pi}{2}}=\frac{-\sin \left(\frac{\pi}{2}\right)}{1+\cos \left(\frac{\pi}{2}\right)}=\frac{-1}{1+0}=-1
\end{aligned}\)
\(\begin{aligned}
& \text { At } \theta=\frac{\pi}{2} \\
& x=\frac{\pi}{2}+\sin \frac{\pi}{2}=1+\frac{\pi}{2}, y=1+\cos \frac{\pi}{2}=1
\end{aligned}\)
\(\therefore \quad\) Equation of the normal is
\(\begin{array}{ll}
& (y-1)=\frac{-1}{\left(\left.\frac{\mathrm{~d} y}{\mathrm{~d} x}\right|_{\theta=\frac{\pi}{2}}\right)}\left(x-1-\frac{\pi}{2}\right) \\
\therefore \quad & (y-1)=1\left(x-1-\frac{\pi}{2}\right) \\
\therefore \quad & 2 x-2 y-\pi=0
\end{array}\)