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
The required value of the limit is 0.