91Ó°ÊÓ

The given equation is \(r = \cot \theta \csc \theta\). We can rewrite this equation in terms of \(\sin \theta\) and \(\cos \theta\) as follows: $$r = \frac{\cos \theta}{\sin \theta} \cdot \frac{1}{\sin \theta}$$ Now, multiply both sides by \(\sin^2 \theta\) to eliminate the denominator: $$r \sin^2 \theta = \cos \theta$$

Now we will substitute the polar to Cartesian conversion formulas (\(x = r \cos \theta\) and \(y = r\sin \theta\)). Recall that $$r^2 = x^2 + y^2,$$ $$x = r \cos \theta,$$ $$y = r \sin \theta.$$ Substituting into our equation, we get: $$\left(\frac{y}{r}\right)^2 = \frac{x}{r}$$ To get rid of the r-terms we can multiply through by \(r^2\): $$y^2 = xr,$$ and substituting \(r = \sqrt{x^2 + y^2}\), we get: $$y^2 = x \sqrt{x^2 + y^2}$$

Now that we have our equation in Cartesian form, let's try to describe the resulting curve. The equation we have is: $$y^2 = x \sqrt{x^2 + y^2}$$ We can see that when \(y=0\), we have \(x=0\), which indicates that the curve must pass through the origin. To further understand the shape of the curve, we can attempt to square both sides of the equation: $$y^4 = x^2(x^2 + y^2)$$ This equation represents a quartic relation between x and y, which implies a higher-order polynomial relation between the coordinates. The exact shape of this curve is not easily determined without graphing. However, it's safe to say that this curve has a complex and non-linear shape that passes through the origin. In conclusion, we have converted the polar equation \(r=\cot \theta \csc \theta\) into Cartesian coordinates as \(y^2 = x \sqrt{x^2 + y^2}\). The resulting curve is a higher-order polynomial relation between x and y that passes through the origin and has a non-linear shape.