Conditions for trigonometric identity $\tan^{-1}x-\tan^{-1}y=\tan^{-1}\dfrac{x-y}{1+xy}$, $x>0$,$y>0$

Question

$\tan^{-1}x-\tan^{-1}y=\tan^{-1}\dfrac{x-y}{1+xy}$, $x>0$,$y>0$

This is the standard trigonometric identity for $x>0$,$y>0$

Now I want to know why it can't be $x>=0,y>=0$

To know that, I tried to prove for $x>=0,y>=0$ just to see if I get any contradiction

$\tan^{-1}x\in\left[0,\dfrac{\pi}{2}\right)$, $\tan^{-1}y\in\left[0,\dfrac{\pi}{2}\right)$

So $\tan^{-1}x-\tan^{-1}y\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ $$A=\tan^{-1}x,B=\tan^{-1}y$$ $$\tan A=x,\tan B=y$$

$$\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$$ $$\tan(A-B)=\dfrac{x-y}{1+xy}$$

Taking $\tan^{-1}$ on both sides

$$\tan^{-1}(\tan(A-B))=\tan^{-1}\dfrac{x-y}{1+xy}$$

As $A-B \in \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$

$$A-B=\tan^{-1}\dfrac{x-y}{1+xy}$$

So I didn't get any contradiction during the entire result. So is $x=0$ or $y=0$ not taken in domain to avoid corner conditions or there is something else?

This is, of course, true for $x=0$. You don't even need to prove it - just substitute $x=0$ into the original formula. Your proof is fine. — Ma Joad, Nov 16 '19 at 03:06
Use $$\tan^{-1}(-z)=-\tan^{-1}z$$See https://math.stackexchange.com/questions/1837410/inverse-trigonometric-function-identity-doubt-tan-1x-tan-1y-pi-tan/1837799#1837799 — lab bhattacharjee, Nov 16 '19 at 05:53
sorry but I didn't get you, $x=0$, $y=0$ is satisfying the original equation. — user3290550, Nov 16 '19 at 05:55
Why do you think that this is invalid for some negative values of $x,y$? Indeed it is, too! The only time it is invalid is when $x,y$ are such that $\arctan x-\arctan y$ falls outside the range $(-π/2,π/2).$ — Allawonder, Nov 16 '19 at 06:27

user3290550 · Answer 1 · 2019-11-16T06:37:35.850

0

Just to add we can't take $x$ as any arbitary R(real), $y$ as any arbitary(real) because of the following issue.

$A=\tan^{-1}x$, $B=\tan^{-1}y$

if $x\in R$, $\tan^{-1}x \in \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$

if $y\in R$, $\tan^{-1}y \in \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$

So $\tan^{-1}x-\tan^{-1}y \in (-\pi,\pi)\tag{1}$

But for $\tan^{-1}x-\tan^{-1}y=\tan^{-1}\dfrac{x-y}{1+xy}$ to be valid, $\tan^{-1}x-\tan^{-1}y \in \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$

But we had $\tan^{-1}x-\tan^{-1}y \in (-\pi,\pi)$ in equation $(1)$

So we can't take $x$ as any arbitary real. Just thought to mention it because one will not find this in many textbooks. May be this would be of some help to beginners.

edited Nov 16 '19 at 06:37

answered Nov 16 '19 at 03:39

user3290550

3,452

The claim that you can't take $x$ or $y$ to be negative is wrong! Take $x=y=-100,$ for example. – Allawonder Nov 16 '19 at 06:29
I am not claiming that x cann't be negative, I am just claiming that we can't allow $x\in R$. Just see the proof again. – user3290550 Nov 16 '19 at 06:30
ok yeah, you were correct, at the last I did mistake, but I have updated my answer now. – user3290550 Nov 16 '19 at 06:32
On the contrary you do. In any case we have to allow $x,y$ to be real, otherwise what do we mean by $\arctan x$ otherwise, for example? – Allawonder Nov 16 '19 at 06:33
bro, I am saying "all" real, – user3290550 Nov 16 '19 at 06:33
Then you'd better clarify that in the earlier parts of your answer. Using the word arbitrary may help. – Allawonder Nov 16 '19 at 06:35
yeah that would be better. – user3290550 Nov 16 '19 at 06:36
so I think the rule of $x>0, y>0$ was made in order to minimize the conditions and to be safe. What do you say? – user3290550 Nov 16 '19 at 06:39
I agree. Although of course that wouldn't necessarily work for the case when you're adding arcs. – Allawonder Nov 16 '19 at 06:41
when we are adding arcs , then also we can't take any arbitary real. There are two conditions in that. – user3290550 Nov 16 '19 at 06:47
Of course we can't. My point was that restricting $x$ and $y$ to be positive wouldn't necessarily help in that case. – Allawonder Nov 16 '19 at 06:59
yeah so we have two cases over there ,$xy<1$ or $xy>1$ and ofcourse $x>0 $ and $y>0$ – user3290550 Nov 16 '19 at 07:05
I disagree with that, but more importantly, I've made the point I wanted to make and really am not interested in bantering further. Thank you. :) – Allawonder Nov 16 '19 at 07:08

Conditions for trigonometric identity $\tan^{-1}x-\tan^{-1}y=\tan^{-1}\dfrac{x-y}{1+xy}$, $x>0$,$y>0$

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