Q.Find the equations of the tangent and normal at $\theta=\dfrac{\pi}{2}$ to the curve $x=a(\theta+\sin\theta)$, $y=a(1+\cos\theta)$.

Find $dy/dx$ in terms of the parameter $\theta$, evaluate at $\theta=\pi/2$, then write the point-slope equations for the tangent and the normal.

1. Differentiate $x$ and $y$ with respect to $\theta$.

   $x=a(\theta+\sin\theta) \Rightarrow \dfrac{dx}{d\theta}=a(1+\cos\theta)$

   $y=a(1+\cos\theta) \Rightarrow \dfrac{dy}{d\theta}=-a\sin\theta$

2. Form $\dfrac{dy}{dx}$.

   $\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}=\dfrac{-a\sin\theta}{a(1+\cos\theta)}=\dfrac{-\sin\theta}{1+\cos\theta}$

3. Evaluate the point at $\theta=\dfrac{\pi}{2}$. Here $\sin\dfrac{\pi}{2}=1$, $\cos\dfrac{\pi}{2}=0$.

   $x=a\left(\dfrac{\pi}{2}+1\right)$, $\quad y=a(1+0)=a$

4. Evaluate the slope at $\theta=\dfrac{\pi}{2}$.

   $\dfrac{dy}{dx}=\dfrac{-1}{1+0}=-1$

5. Equation of the tangent (slope $m=-1$ through $\left(a\left(\dfrac{\pi}{2}+1\right),\,a\right)$):

   $y-a=-1\left[x-a\left(\dfrac{\pi}{2}+1\right)\right]$

   $y-a=-x+a\left(\dfrac{\pi}{2}+1\right)$

   …