# What is the Cartesian form of #r^2sintheta = 2theta-4tantheta-csctheta #?

Please see the explanation

Allowing y to equal zero would be trouble.

Allowing x to equal zero is a division by zero issue.

I am saving the best for last so let's write the equation with these 3 substitutions:

This creates the 3 following equations:

Undefined elsewhere.

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To convert the polar equation (r^2 \sin(\theta) = 2\theta - 4\tan(\theta) - \csc(\theta)) to Cartesian form, we can use the relationships between polar and Cartesian coordinates:

[x = r \cos(\theta)] [y = r \sin(\theta)]

First, we need to express (r) in terms of (x) and (y). Since (r^2 = x^2 + y^2), we have (r = \sqrt{x^2 + y^2}).

Now, substitute (r) in the given equation with (\sqrt{x^2 + y^2}):

[\sqrt{x^2 + y^2} \cdot \sin(\theta) = 2\theta - 4\tan(\theta) - \csc(\theta)]

Then, replace (\sin(\theta)) and (\csc(\theta)) with their Cartesian equivalents:

[\frac{y}{\sqrt{x^2 + y^2}} = 2\theta - 4\left(\frac{y}{x}\right) - \frac{1}{\sin(\theta)}]

Since (\sin(\theta) = \frac{y}{r}) and (r = \sqrt{x^2 + y^2}), we can substitute (\sin(\theta)) with (\frac{y}{\sqrt{x^2 + y^2}}):

[\frac{y}{\sqrt{x^2 + y^2}} = 2\theta - 4\left(\frac{y}{x}\right) - \frac{\sqrt{x^2 + y^2}}{y}]

To eliminate (\theta), we need to express it in terms of (x) and (y). We know that (\tan(\theta) = \frac{y}{x}), so (\theta = \arctan\left(\frac{y}{x}\right)).

Substitute (\theta) with (\arctan\left(\frac{y}{x}\right)):

[\frac{y}{\sqrt{x^2 + y^2}} = 2\arctan\left(\frac{y}{x}\right) - 4\left(\frac{y}{x}\right) - \frac{\sqrt{x^2 + y^2}}{y}]

This is the Cartesian form of the given polar equation.

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