x=cos^2theta ,y=costheta Find all points (if any) of horizontal and vertical tangency to the curve.

x=cos^2 theta
y=cos theta
First, take the derivative of x and y with respect to theta .
dx/(d theta) = 2costheta (-sin theta)
dx/(d theta)=-2sintheta cos theta
dy/(d theta) = -sin theta
Take note that the slope of a 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 is zero.
0= (dy/(d theta))/(dx/(d theta))
This implies that the graph of the parametric equation will have a horizontal tangent when dy/(d theta)=0 and dx/(d theta)!=0 .
So, set the derivative of y equal to zero.
dy/(d theta) = 0
-sin theta = 0
sin theta=0
theta= 0, pi
Then, plug-in these values of theta to dx/(d theta) to verify if the slope is zero, not indeterminate.
dx/(d theta) =-2sintheta cos theta
theta_1=0
dx/(d theta) =-2sin(0) cos (0) = 0
theta_2=pi
dx/(d theta) = -2sin(pi)cos(pi)=0
Since both dy/(d theta) and dx/(d theta) are zero at these values of theta, the slope is indeterminate.
m=dy/dx= (dy/(d theta))/(dx/(d theta))=0/0  (indeterminate)
Therefore, the parametric equation has no horizontal tangent.
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 happens when dx/(d theta)=0 , but dy/(d theta)!=0 .
So, set the derivative of x equal to zero.
dx/(d theta) = 0
-2sinthetacos theta=0
-2sin theta = 0
sin theta = 0
theta = 0, pi
cos theta = 0
theta = pi/2, (3pi)/2
Take note that at theta =0 and theta =pi , both dy/(d theta) and dx/(d theta) are zero.  So the slope at these two values of theta is indeterminate.
m=dy/dx= (dy/(d theta))/(dx/(d theta))=0/0  (inderterminate)
Plug-in theta =pi/2 and theta=(3pi)/2 to dy/(d theta) to verify that the slope is undefined at these values of theta.
dy/(d theta) = -sin theta
theta_1=pi/2
dy/(d theta) = -sin (pi/2)=-1
theta_2= (3pi)/2
dy/(d theta) = -sin ((3pi)/2)=1
Since  dy/(d theta) !=0 ,  the parametric equation has vertical tangent at  theta_1=pi/2 and theta=(3pi)/2 .
Then, plug-in these values to the parametric equation to get the points (x,y).
x=cos^2 theta
y=cos theta
theta_1= pi/2
x=cos^2(pi/2)=0
y= cos (pi/2)=0
theta_2= (3pi)/2
x=cos^((3pi)/2)=0
y=cos((3pi)/2)=0
Since theta_1 and theta_2 result to same x and y coordinates, there is only one point in which the curve has a vertical tangent.
Therefore, the graph of the parametric equation has vertical tangent at point (0,0).