Calculus: Early Transcendentals, Chapter 4, 4.4, Section 4.4, Problem 11

You need to evaluate the limit, hence, you need to replace pi/2 for x such that:
lim_(x->pi/2) (cos x)/(1 - sin x) = (cos (pi/2))/(1 - sin (pi/2)) = 0/(1-1) = 0/0
Since the limit is indeterminate 0/0, you may apply l'Hospital's rule:
lim_(x->pi/2) (cos x)/(1 - sin x) = lim_(x->pi/2) ((cos x)')/((1 - sin x)')
lim_(x->pi/2) ((cos x)')/((1 - sin x)')= lim_(x->pi/2) (-sin x)/(-cos x)
Replacing by pi/2 yields:
lim_(x->pi/2) (-sin x)/(-cos x) = lim_(x->pi/2) tan x = tan (pi/2)
If x->pi/2 and xIf x->pi/2 and x>pi/2 , then tan (pi/2) = -oo
Hence, evaluating the limit, yields lim_(x->pi/2) (cos x)/(1 - sin x) = +oo o if x->pi/2 and xpi/2) (cos x)/(1 - sin x) = -oo , if x->pi/2 and x>pi/2.