Precalculus, Chapter 4, 4.2, Section 4.2, Problem 53

True
tan(a) = tan(a - 6pi)
proof
tan(a - 6pi) = sin(a - 6pi)/cos(a - 6pi)
Now,
sin(a - 6pi) = ?
this is of the form
sin(a-b) = sin(a)cos(b) - cos(a)sin(b)
so ,
sin(a - 6pi) =sin(a)cos(6pi) - cos(a)sin(6pi)
= sin(a) (1) -0 = sin(a)

cos(a - 6pi) = ?
this is of the form
cos(a-b)= cos(a)cos(b) +sin(a)sin(b)
so,
cos(a - 6pi) = cos(a)cos(6pi) +sin(a)sin(6pi)
= cos(a) (1) + 0 = cos(a)
so , now
tan(a - 6pi) = sin(a - 6pi)/cos(a - 6pi) = sin(a)/cos(a) = tan(a)