Calculus of a Single Variable, Chapter 10, 10.3, Section 10.3, Problem 37

x=sec theta
y=tan theta
First, take the derivative of x and y with respect to theta.
dx/(d theta) = sec theta tan theta
dy/(d theta) = sec^2 theta
Then, set each derivative equal to zero.
dx/(d theta) =0
sec theta tan theta=0

sec theta = 0
theta={O/} (There are no angles in which secant is zero.)
tan theta =0
theta = 0, pi

dy/ (d theta ) = 0
sec^2 theta=0

sec theta =0
theta = {O/} (There are no angles in which secant is zero.)

Take note that the slope of a tangent is equal to dy/dx.
m =dy/dx
To get 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 .
For the given parametric equation, there are no values of theta in which the dy/(d theta) is zero.
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 implies that the slope is undefined when dx/(d theta)=0 , but dy/(d theta)!=0 . This occurs at theta =0 and theta=pi .
Then, plug-in these values to the parametric equation to get the points (x,y).
theta_1=0
x_1=sec (0)=1
y_1=tan(0)=0
theta_2=pi
x_2=sec (pi)=-1
y_2=tan(pi)=0
Therefore, the parametric equation has vertical tangent at point (1,0) and (-1,0).