We know that as #cos(x+y)=cosxcosy-sinxsiny# and #cos(x-y)=cosxcosy+sinxsiny#, hence
#cos(x+y)/cos(x-y)=(cosxcosy-sinxsiny)/(cosxcosy+sinxsiny)#
Now applying componendo-dividendo (for details see note below), we get
#(cos(x+y)+cos(x-y))/(cos(x+y)-cos(x-y))=(cosxcosy-sinxsiny+cosxcosy+sinxsiny)/(cosxcosy-sinxsiny-cosxcosy-sinxsiny)#
= #(2cosxcosy)/(-2sinxsiny)#
= #-(cosxcosy)/(sinxsiny)#
or #(cos(x+y)+cos(x-y))/(cos(x-y)-cos(x+y))=(cosxcosy)/(sinxsiny)#
What is componendo-dividendo #-# If #a/b=c/d#, then adding #1# to each side, we get #a/b+1=c/d+1# or #(a+b)/b=(c+d)/d#. Similarly subtracting #1#, we get #a/b-1=c/d-1# or #(a-b)/b=(c-d)/d#. Now dividing #(a+b)/b=(c+d)/d# by #(a-b)/b=(c-d)/d#, we get
#(a+b)/(a-b)=(c+d)/(c-d)#. Hence, if #a/b=c/d#, then #(a+b)/(a-b)=(c+d)/(c-d)#
