What is the value of lim x->pi/2 [(1-sinx)/cos x]

We have to find the value of lim x-->pi/2 [(1 - sin
x)/cos x]

Substituting x = pi/2, gives us 0/0, which is
indeterminate. This allows us to use the L'Hopital's rule and substitute the numerator
and denominator with their derivatives

=> lim
x-->pi/2 [(-cos x)/-sin x]

substituting x =
pi/2

=>
0

The required value of the limit is
0.