How do you simplify the expression #secxcotx-cotxcosx#?
1 Answer
Aug 28, 2016
Explanation:
We will use the following:
#secx = 1/cosx# #cotx = cosx/sinx# #sin^2x + cos^2x = 1 => 1 - cos^2x = sin^2x#
#=1/sinx-cos^2x/sinx# (assuming#cosx != 0# )
#=(1-cos^2x)/sinx#
#=sin^2x/sinx#
#=sinx# (assuming#sinx != 0# )