A question about the arctangent addition formula.

In the arctangent formula, we have that:

arctanu+arctanv=arctan(u+v1uv)

however, only for uv<1. My question is: where does this condition come from? The situation is obvious for uv=1, but why the inequality?

One of the possibilities I considered was as following: the arctangent addition formula is derived from the formula:

tan(α+β)=tanα+tanβ1tanαtanβ.

Hence, if we put u=tanα and v=tanβ (which we do in order to obtain the arctangent addition formula from the one above), the condition that uv<1 would mean tanαtanβ<1, which, in turn, would imply (thought I am NOT sure about this), that π/2<α+β<π/2, i.e. that we have to stay in the same period of tangent.

However, even if the above were true, I still do not see why we have to stay in the same period of tangent for the formula for tan(α+β) to hold. I would be thankful for a thorough explanation.

Answer

If arctanx=A,arctany=B; tanA=x,tanB=y

We know, tan(A+B)=tanA+tanB1tanAtanB

So, tan(A+B)=x+y1xy
arctan(x+y1xy)=nπ+A+B=nπ+arctanx+arctany where n is any integer

As the principal value of arctanz lies [π2,π2],πarctanx+arctanyπ

(1) If π2<arctanx+arctanyπ,arctan(x+y1xy)=arctanx+arctanyπ to keep arctan(x+y1xy)[π2,π2]

Observe that arctanx+arctany>π2arctanx,arctany>0x,y>0

arctanx>π2arctany
x>tan(π2arctany)=cotarctany=cot(arccot1y)x>1yxy>1

(2) If πarctanx+arctany<π2,arctan(x+y1xy)=arctanx+arctany+π

Observe that arctanx+arctany<π2arctanx,arctany<0x,y<0

Let x=X2,y=Y2

arctan(X2)+arctan(Y2)<π2
arctan(X2)<π2arctan(Y2)
X2<tan(π2arctan(Y2))=cotarctan(Y2)=cot(arccot1Y2)

X2<1Y2X2>1Y2X2Y2>1xy>1

(3) If π2arctanx+arctanyπ2,arctanx+arctany=arctan(x+y1xy)

Attribution
Source : Link , Question Author : Johnny Westerling , Answer Author : lab bhattacharjee

Leave a Comment