How do you simplify #(1+sin y)/( 1+csc y)#?

Dec 2, 2015

#siny#

#(1+siny)/(1+cscy)#

#=(1+siny)/(1+1/siny)#

#=(1+siny)/((siny+1)/siny)#

#=(1+siny)*(siny/(siny+1))#

#=color(red)cancelcolor(black)(1+siny)*(siny)/color(red)cancelcolor(black)(siny+1)#

#=siny#

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