In the arctangent formula, we have that:

arctanu+arctanv=arctan(u+v1−uv)

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+tanβ1−tanαtanβ.

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+tanB1−tanAtanB

So, tan(A+B)=x+y1−xy

⟹arctan(x+y1−xy)=nπ+A+B=nπ+arctanx+arctany where n is any integer

As the principal value of arctanz lies ∈[−π2,π2],−π≤arctanx+arctany≤π

(1) If π2<arctanx+arctany≤π,arctan(x+y1−xy)=arctanx+arctany−π to keep arctan(x+y1−xy)∈[−π2,π2]

Observe that arctanx+arctany>π2⟹arctanx,arctany>0⟹x,y>0

⟹arctanx>π2−arctany

⟹x>tan(π2−arctany)=cotarctany=cot(arccot1y)⟹x>1y⟹xy>1

(2) If −π≤arctanx+arctany<−π2,arctan(x+y1−xy)=arctanx+arctany+π

Observe that arctanx+arctany<−π2⟹arctanx,arctany<0⟹x,y<0

Let x=−X2,y=−Y2

⟹arctan(−X2)+arctan(−Y2)<−π2

⟹arctan(−X2)<−π2−arctan(−Y2)

⟹−X2<tan(−π2−arctan(−Y2))=cotarctan(−Y2)=cot(arccot−1Y2)

⟹−X2<1−Y2⟹X2>1Y2⟹X2Y2>1⟹xy>1

(3) If −π2≤arctanx+arctany≤π2,arctanx+arctany=arctan(x+y1−xy)

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