$xy<1 \iff \text{arctan }x + \text{arctan }y \in (-\pi/2,\pi/2)$

Question

Is this claim true?

$$xy<1 \iff \text{arctan }x + \text{arctan }y \in (-\pi/2,\pi/2)$$

If so, how to prove?

I was led to this while trying to figure out the Addition Formula for Arctangent. I've looked at many questions and answers about that Formula but don't seem to have come across a proof of the above claim.

https://math.stackexchange.com/questions/1837410/inverse-trigonometric-function-identity-doubt-tan-1x-tan-1y-pi-tan — lab bhattacharjee, Mar 24 '20 at 06:23
It seems that it is included in the question https://math.stackexchange.com/questions/1724348/what-is-arctanx-arctany — user159888, Mar 24 '20 at 06:23

score 2 · Answer 1 · answered Mar 24 '20 at 06:25

It is enough to prove that $\arctan x+\arctan y <\pi/2$ since we can change $x$ to $-x$ and $y $ to $-y$ to get the lower bound.

Let $y >0$. Now $\arctan x+\arctan y$ is a strictly increasing function of $x$ so it is enough to prove that $\arctan y+\arctan \frac 1y \leq \pi /2$. You can check that the derivative of the left side is negative for $y <1$ positive for $y>1$. Hence the minimum value is attained when $y=1$. But the value when $y=1$ is $\frac {\pi} 4+\frac {\pi} 4=\frac {\pi} 2$.

I will leave the case $y <0$ to you.

Yes, you were right, thanks! – K.K.McDonald Mar 24 '20 at 06:28 — K.K.McDonald, Mar 24 '20 at 06:28

score 1 · Accepted Answer · answered Mar 24 '20 at 07:20

Below, I use these two results:

$\tan^{-1}$ is strictly increasing.
$$\tan^{-1}x+\tan^{-1}\frac{1}{x}=\begin{cases} \pi/2, & \text{for }x>0,\\ -\pi/2, & \text{for }x<0. \end{cases}$$

Note that if $x=0$, then both conditions hold. So below I'll assume $x\neq0$.

($\implies$) Suppose $xy<1$.

Case 1. $x>0$.

Since $y<1/x$ and $\tan^{-1}$ is strictly increasing, $$\tan^{-1}x+\tan^{-1}y<\tan^{-1}x+\tan^{-1}\frac{1}{x}=\frac{\pi}{2}.$$

Since $\tan^{-1}x>0$ and $\tan^{-1}y>-\pi/2$, $$\tan^{-1}x+\tan^{-1}y>-\frac{\pi}{2}.$$

Case 2. $x<0$. (Similar, omitted).

($\impliedby$) Suppose $\tan^{-1}x+\tan^{-1}y \in (-\pi/2,\pi/2)$.

Case 1. $x>0$.

$$\tan^{-1}x+\tan^{-1}y<\frac{\pi}{2}=\tan^{-1}x+\tan^{-1}\frac{1}{x}.$$

Since $\tan^{-1}$ is strictly increasing, $y<1/x$ or equivalently $xy<1$.

Case 2. $x<0$. (Similar, omitted).

score 1 · Answer 3 · answered Mar 24 '20 at 08:22

Let $\;\alpha = \text{Arc}\tan x, \;\beta = \text{Arc}\tan y \;\Rightarrow\; \alpha,\beta \;\in \;(-\pi/2, \pi/2).$

The assertion is that $\;xy < 1 \;\Leftrightarrow (\alpha + \beta) \;\in \;(-\pi/2, \pi/2).$

$\Rightarrow$
Given that $\;xy < 1.$
WLOG $\;\alpha, \beta\;$ have the same sign, else it is immediate that $(\alpha + \beta) \;\in \;(-\pi/2, \pi/2).$

$\alpha, \beta > 0\;$ and $\;(xy) < 1 \Rightarrow$
$0 < x, \;0 < y,\;$ and $\;y < (1/x) \;\Rightarrow$
$\tan \beta \;<\; \frac{1}{\tan \alpha} = \cot \alpha = \tan (\pi/2 \;-\; \alpha) \;\Rightarrow$
$\beta < (\pi/2 \;-\; \alpha) \Rightarrow (\alpha + \beta) \;\in \;(0, \pi/2).$

$\alpha, \beta < 0\;$ and $\;(xy) < 1 \Rightarrow$
$x < 0, \;y < 0,\;$ and $\;[$since $x < 0]\;$ $\;y > (1/x) \;\Rightarrow$
$\tan \beta \;>\; \frac{1}{\tan \alpha} = \cot \alpha = \tan (-\pi/2 \;-\; \alpha) \;\Rightarrow$
$\beta > (-\pi/2 \;-\; \alpha) \Rightarrow (\alpha + \beta) \;\in \;(-\pi/2, 0).$

$\Leftarrow$
Given that $\;(\alpha + \beta) \;\in \;(-\pi/2, \pi/2).$
If $\alpha$ and $\beta$ have different signs
then $x$ and $y$ have different signs $\;\Rightarrow\; xy < 0 < 1.$
WLOG, $\alpha$ and $\beta$ have the same sign.

$0<\alpha, 0<\beta\;$ and $\;(\alpha + \beta) < \pi/2 \;\Rightarrow$
$[\;0 < x, 0 < y\;$ and $\;\beta < (\pi/2 - \alpha)\;] \;\Rightarrow$
$y = \tan \beta < \tan (\pi/2 - \alpha) = \cot \alpha = (1/x) \;\Rightarrow xy < 1.$

$\alpha<0, \beta<0\;$ and $\;(\alpha + \beta) > -\pi/2 \;\Rightarrow$
$[\;x < 0, y < 0\;$ and $\;\beta > (-\pi/2 - \alpha)\;] \;\Rightarrow$

$y = \tan \beta > \tan (-\pi/2 - \alpha) = \cot \alpha = (1/x) \;\Rightarrow $
[since $x < 0]\; xy < 1.$

$xy<1 \iff \text{arctan }x + \text{arctan }y \in (-\pi/2,\pi/2)$

3 Answers3