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

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