2tan-1(-3)=-π/2+tan-1(-4/3).
Let a=tan-1(-3), so tan(a)=-3; and b=tan-1(-4/3), so tan(b)=-4/3.
b-π/2=2a;
tan(b-π/2)=-tan(π/2-b)=-cot(b)=-1/tan(b)=¾.
tan(2a)=2tan(a)/(1-tan2(a))=-6/(1-9)=-6/-8=¾.
So tan(2a)=tan(b-π/2), making 2a=-π/2+b.
Therefore 2a=-π/2+b and, substituting for a and b, 2tan-1(-3)=-π/2+tan-1(-4/3) QED