How do you convert $r=\dfrac{1}{1-\cos \theta }$ in rectangular form?

Hint: In the above problem, we have given the polar form and asked us to convert into rectangular form. Polar form contains $\left( r,\theta \right)$ form and the rectangular form contains $\left( x,y \right)$. We know the conversion of $\left( r,\theta \right)$ into $\left( x,y \right)$ as: $x=r\cos \theta ,y=r\sin \theta $ so using these conversions, we can convert the given polar form into rectangular form.

Complete step by step solution:
The polar form given in the above problem is as follows:
$r=\dfrac{1}{1-\cos \theta }$
Now, cross – multiplying in the above equation we get,
$\Rightarrow r\left( 1-\cos \theta \right)=1$
Multiplying r inside the bracket we get,
$\Rightarrow r-r\cos \theta =1............(1)$
We know the conversions from polar form to rectangular form as follows:
$\begin{align}
  & x=r\cos \theta ; \\
 & y=r\sin \theta \\
\end{align}$
Also, we know the relation between x, y and r as follows:
${{x}^{2}}+{{y}^{2}}={{r}^{2}}$
Taking square root in the above equation we get,
$\Rightarrow \sqrt{{{x}^{2}}+{{y}^{2}}}=r$
Substituting the value of $r\And r\cos \theta $ in eq. (1) we get,
$\Rightarrow \sqrt{{{x}^{2}}+{{y}^{2}}}-x=1$
Adding x on both the sides of the above equation we get,
$\Rightarrow \sqrt{{{x}^{2}}+{{y}^{2}}}=1+x$
Squaring on both the sides of the above equation we get,
$\begin{align}
  & \Rightarrow {{\left( \sqrt{{{x}^{2}}+{{y}^{2}}} \right)}^{2}}={{\left( 1+x \right)}^{2}} \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}=1+{{x}^{2}}+2x \\
\end{align}$
In the above equation, ${{x}^{2}}$ will be cancelled out on both the sides of the equation we get,
${{y}^{2}}=1+2x$
Hence, we have converted the given polar form into rectangular form as follows:
${{y}^{2}}=1+2x$

Note: In the above solution, we have written a relation between x, y and r as follows:
${{x}^{2}}+{{y}^{2}}={{r}^{2}}$
The proof of the above equation is that in the above solution we have written the relation between x, r and $\cos \theta $. Also, we have written the relation between y, r and $\sin \theta $.
$\begin{align}
  & x=r\cos \theta .....(2) \\
 & y=r\sin \theta .......(3) \\
\end{align}$
Squaring both the equations and then adding them we get,
$\begin{align}
  & {{x}^{2}}={{r}^{2}}{{\cos }^{2}}\theta \\
 & {{y}^{2}}={{r}^{2}}{{\sin }^{2}}\theta \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}={{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right).........(4) \\
\end{align}$
In the above equation, we can use the following trigonometric identity which is equal to
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
Using the above relation in eq. (4) we get,
$\begin{align}
  & \Rightarrow {{x}^{2}}+{{y}^{2}}={{r}^{2}}\left( 1 \right) \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}={{r}^{2}} \\
\end{align}$