Find the equation

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

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

$\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a}(\sin \theta)$

Now dividing $\frac{d y}{d \theta}$ and $\frac{d x}{d \theta}$ to obtain the slope of tangent

$\frac{d y}{d x}=\frac{\sin \theta}{1+\cos \theta}$

$\mathrm{m}$ (tangent) at theta is $\frac{\sin \theta}{1+\cos \theta}$

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

$\mathrm{m}$ (normal) at theta is $-\frac{\sin \theta}{1+\cos \theta}$

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

$y-a(1-\cos \theta)=\frac{\sin \theta}{1+\cos \theta}(x-a(\theta+\sin \theta))$

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

$y-a(1-\cos \theta)=\frac{1+\cos \theta}{-\sin \theta}(x-a(\theta+\sin \theta))$