How do you simplify #cotx / (cscx)#?

1 Answer
Dec 5, 2015

#cosx#

Explanation:

Recall:
#1. cotx=1/tanx# or #cosx/sinx#
#2. cscx=1/sinx#

Substitute your reciprocal and quotient identities into the equation:

#cotx/cscx#

#=(cosx/sinx)/(1/sinx)lArr# use #cotx=cosx/sinx# instead of #1/tanx#

#=cosx/sinx-:1/sinx#

#=cosx/sinx*sinx/1#

#=cosx/color(red)cancelcolor(black)sinx*color(red)cancelcolor(black)sinx/1#

#=cosx#