Find the equation

finding slope of the tangent by differentiating $x$ and $y$ with respect to theta

$\frac{\mathrm{dx}}{\mathrm{d} \theta}=1+\cos \theta$

$\frac{d y}{d \theta}=-\sin \theta$

Dividing both the above equations

$\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\sin \theta}{1+\cos \theta}$

$\mathrm{m}$ (tangent) at theta $(\pi / 2)=-1$

normal is perpendicular to tangent so, $m_{1} m_{2}=-1$

$m$ (normal) at theta $(\pi / 2)=1$

equation of tangent is given by $y-y_{1}=m($ tangent $)\left(x-x_{1}\right)$

$y-1=-1\left(x-\frac{\pi}{2}-1\right)$

equation of normal is given by $y-y_{1}=m($ normal $)\left(x-x_{1}\right)$

$y-1=1\left(x-\frac{\pi}{2}-1\right)$