tan-1(2x) = tan-1(2x1-x2) -1<x<1
= pi + tan-1(2x1-x2) x>1
= -pi + tan-1(2x1-x2) x<-1
So putting tanx in place of x
we have tan-1(2tanx) = 2x, -1<tanx<1
= pi+2x tanx>1
= -pi + 2x tanx < -1
tan-1(2x) = tan-1(2x1-x2) -1<x<1
= pi + tan-1(2x1-x2) x>1
= -pi + tan-1(2x1-x2) x<-1
So putting tanx in place of x
we have tan-1(2tanx) = 2x, -1<tanx<1
= pi+2x tanx>1
= -pi + 2x tanx < -1