`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 tangent when `dy/(d theta)=0` and `dx/(d theta) !=0`. 

Setting the derivative of y yields:

`dy/(d theta) = 0`

`costheta - 2sinthetacostheta=0`

`costheta(1 - 2sintheta) =0`

`costheta = 0`

`theta=pi/2,(3pi)/2`

`1-2sintheta=0`

`-2sintheta=-1`

`sintheta=1/2`

`theta=pi/6,(5pi)/6`

Take note that at `theta=pi/2` , the value of `dx/(d theta)` is zero. Since both `dy/(d theta)` and `dx/(d theta)`  are zero, the slope at this value of theta is indeterminate.

`m=0/0`   (indeterminate)

So the polar curve has horizontal tangents at:

`theta_1 = pi/6 + 2pin`

`theta_2= (5pi)/6+2pin`

`theta_3= (3pi)/2+2pin`

where n is any integer.

To determine the points `(r, theta)` , plug-in the values of theta to the polar equation.

`r=1-sin theta`

`theta_1 = pi/6 + 2pin`

`r_1=1-sin(pi/6 + 2pin)=1-sin(pi/6) = 1-1/2=1/2`

`theta_2= (5pi)/6+2pin`

`r_2=1-sin((5pi)/6+2pin)=1-sin((5pi)/6)= 1 -1/2=1/2`

`theta_3= (3pi)/2+2pin`

`r_3=1-sin((3pi)/2+2pin)=1-sin((3pi)/2)=1-(-1)=2`

Therefore, the polar curve has horizontal tangent at points

`(1/2, pi/6+2pin)` ,   `(1/2, (5pi)/6+2pin)` ,  and  `(2, (3pi)/2+2pin)` .

Moreover, when the tangent line is vertical, the slope is undefined.

`u n d e f i n e d =(dy/(d theta)) / (dx/(d theta))`

This implies that the polar curve will have vertical tangent when `dx/(d theta)=0` and `dy/(d theta)!=0` .

Setting the derivative of x equal to zero yields:

`dx/(d theta) = 0`

`-sintheta+sin^2theta-cos^2theta=0`

`-sin theta + sin^2 theta-(1-sin^2 theta) = 0`

`2sin^2 theta -sin theta -1=0`

`(2sin theta +1)(sin theta -1) = 0`

`2sin theta + 1=0`

`sin theta=-1/2`

`theta = (7pi)/6,(11pi)/6`

`sin theta -1=0`

`sin theta=1`

`theta=pi/2`

Take note that at `theta =pi/2` , both `dy/(d theta )` and `dx/(d theta)` are zero. So the slope is indeterminate at this value of theta.

`m=0/0`  (indeterminate)

So the polar curve has vertical tangents at:

`theta_1 =(7pi)/6+2pin`

`theta_2=(11pi)/6+2pin`

where n is any integer.

To determine the points `(r, theta)` , plug-in the values of theta to the polar equation.

`r=1-sin theta`

`theta_1=(7pi)/6+2pin`

`r_1=1-sin((7pi)/6+2pin)=1-sin((7pi)/6)=1-(-1/2)=3/2`

`theta_2=(11pi)/6+2pin`

`r_2=1-sin((11pi)/6+2pin)=1-sin((11pi)/6)=1-(-1/2)=3/2`

Therefore, the polar curve has vertical tangent at points `(3/2, (7pi)/6+2pin)` and `(3/2, (11pi)/6+2pin)` .

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