posted by Jay .
True or false?
tan x = tan(x-6pi)
Using tan(A-B) = (tanA - tanB)/(1+tanAtanB)
RS = tan(x-6π)
= (tanx - tan 6π)/(1 + tanxtan 6π)
but 6π is coterminal with 2π (6π is 3 rotations, and 2π is one rotation
so tan 6π = tan 2π = 0
= (tanx - 0)/(1+ ) = tanx = LS
so the statement is true
tan(-A) = -tanA
so tan(x-6π) = - tan(6π-x)
= -tan(2π-x) , since 6π is coterminal with 2π
= -( -tan(x) ) , by the CAST rule