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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
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