Answer :
Recall the following rules for identifying the rectangular coordinates (x,y) from the polar coordinates (r,θ):
[tex]\begin{gathered} x=r\cos\theta \\ y=r\sin\theta \end{gathered}[/tex][tex]r=\sqrt{x^2+y^2}[/tex]First, rewrite cot(2θ)csc(θ) in terms of sines and cosines:
[tex]\begin{gathered} r=(\cot2\theta)(\csc\theta) \\ \\ =\frac{\cos2\theta}{\sin2\theta}\cdot\frac{1}{\sin\theta} \\ \\ =\frac{1-2\sin^2\theta}{2\sin\theta\cos\theta}\cdot\frac{1}{\sin\theta} \\ \\ =\frac{1-2\sin^2\theta}{2\sin^2\theta\cos\theta} \end{gathered}[/tex]Multiply both numerator and denominator by r^3 and order the factors in such a way that the expressions for x and y are made evident:
[tex]\begin{gathered} r=\frac{1-2\sin^2\theta}{2\sin^2\theta\cos\theta} \\ \\ \Rightarrow r=\frac{r^3(1-2\sin^2\theta)}{r^3(2\sin^2\theta\cos\theta)} \\ \\ =\frac{r(r^2-2(r^2\sin^2\theta))}{2(r^2\sin^2\theta)(r\cos\theta)} \\ \\ =\frac{r(r^2-2(r\sin\theta)^2)}{2(r\sin\theta)^2(r\cos\theta)} \end{gathered}[/tex]Simplify and replace the expressions for rsin(θ) and rcos(θ) as well as r:
[tex]\begin{gathered} r=\frac{r(r^2-2(r\sin\theta)^2)}{2(r\sin\theta)^2(r\cos\theta)} \\ \\ \Rightarrow1=\frac{(r^2-2(r\sin\theta)^2)}{2(r\sin\theta)^2(r\cos\theta)} \\ \\ \Rightarrow1=\frac{(x^2+y^2-2(y)^2)}{2(y)^2(x)} \\ \\ \Rightarrow1=\frac{(x^2+y^2-2y^2)}{2y^2x} \\ \\ \Rightarrow1=\frac{(x^2-y^2)}{2y^2x} \\ \\ \Rightarrow2y^2x=x^2-y^2 \end{gathered}[/tex]Isolate y^2 from the equation:
[tex]\begin{gathered} \Rightarrow2y^2x+y^2=x^2 \\ \\ \Rightarrow y^2(2x+1)=x^2 \\ \\ \Rightarrow y^2=\frac{x^2}{2x+1} \end{gathered}[/tex]Therefore, the polar equation can be written as a rectangular equation as follows:
[tex]y^2=\frac{x^2}{2x+1}[/tex]