Question

True or false?

tan x = tan(x-6pi)

Answer 1

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

or

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
= tanx

Answer 2

the statement is true
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 tru

Answer 3

Sona really just copied Reiny like that