Question #465f7
1 Answer
Oct 13, 2016
Explanation:
#= (1-cos^2x)/(1-cosx)#
# = ((1-cosx)(1+cosx))/(1-cosx)#
# = 1+cosx# .
So,
#f'(x) = -sinx# .
#= (1-cos^2x)/(1-cosx)#
# = ((1-cosx)(1+cosx))/(1-cosx)#
# = 1+cosx# .
So,
#f'(x) = -sinx# .