`r=1-sin theta`

To solve, express the polar equation in parametric form. To convert it to parametric equation, apply the formula

`x = rcos theta`

`y=r sin theta`

Plugging in `r=1-sin theta` , the formula becomes:

`x=(1-sin theta)cos theta=cos theta -sin theta cos theta`

`y = (1-sin theta)sin theta=sin theta -sin^2 theta`

So the equivalent parametric equation of `r= 1-sin theta` is:

`x=cos theta -sin theta cos theta`

`y=sin theta -sin^2 theta`

Then, take the derivative of x and y with respect to theta.

`dx/(d theta) = -sintheta - (sintheta*(-sintheta) + costheta*costheta)`

`dx/(d theta)=-sintheta+sin^2theta-cos^2theta`

`dy/(d theta) = costheta - 2sinthetacostheta`

Take note that the slope of the tangent is equal to dy/dx.

`m= (dy)/(dx)`

To get the dy/dx of a parametric equation, apply the formula:

`dy/dx = (dy/(d theta))/(dx/(d theta))`

When the tangent line is horizontal, the slope of the tangent is zero.

`0 = (dy/(d theta)) / (dx/(d theta))`

This implies that the polar curve will have a horizontal...

(The entire section contains 529 words.)

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