Calculus of a Single Variable, Chapter 5, 5.3, Section 5.3, Problem 34

f(x)=sec(x)
Take note that a function is strictly monotonic on a given interval if it is entirely increasing on that interval or entirely decreasing on that interval.
To determine if f(x) is strictly monotonic on the interval [0, pi/2) , let's take its derivative.
f(x)=sec(x)
f'(x) =sec(x)tan(x)
Then, determine the critical numbers. To do so, set f'(x) equal to zero.
0=sec(x)tan(x)
Then, set each factor equal to zero
secx=0
x= {O/ }
(There are no angles in which the value of secant will be zero.)
tanx=0
x={0,pi,2pi,...pik}
So on the interval [0,pi/2) , the only critical number that belongs to it is x=0. Since the critical number is the boundary of the given interval, it indicates that the there is no sign change in the value of f'(x) on [0, pi/2). To verify, let's assign values to x which falls on that interval and plug-in them to f'(x).
f'(x) = sec(x)tan(x)

x=pi/6
f'(x)=sec(pi/6)tan(pi/6)=(2sqrt3)/3*sqrt3/3=(2*3)/3=2/3
x=pi/4
f'(x)=sec(pi/4)tan(pi/4)=sqrt2*1=sqrt2
x=pi/3
f'(x)=sec(pi/3)tan(pi/3)=2*sqrt3=2sqrt3
Notice that on the interval [0, pi/2) , the values of f'(x) are all positive. There is no sign change. So the function is entirely increasing on this interval.
Therefore, the function f(x)=sec(x) is strictly monotonic on the interval [0,pi/2) .