If (sin (x+y))÷(cos (x-y))=(1-m)÷(1+m), then prove that tan (45-x)×tan (45-y) = m

sin(x+y) /Cos(x-y) =(1-m)/(1+m)
so 
(sinxcosy+cosxsiny)/(sinxsiny-cosxcosy) =(1-m)/(1+m)
So divide by cosxcosy in LHS 
(tanx+tany)/(tanxtany-1)=(1-m)/(1+m)
tan(x+y)=(m-1)/(m+1)
componeneto 

(tan(x+y)+1)/(tan(x+y)-1)=(2m)/(-2)

tan (45-x)×tan (45-y) = m
{(1-tanx)/(1+tanx)}{(1-tany)/(1+tany)}=m
{(1+tanxtay-tanx-tany)/{(1+tanxtany+tanx+tany)}=m