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I am trying to understand how the inverse trigonometric identities below work/are derived: $$\tan^{-1}(x)+\tan^{-1}(y)=\tan^{-1}\left(\frac{x+y}{1-xy}\right)\\ xy<1$$ The above identity has a proof that depends on the identity for $\tan(x)+\tan(y)$, which I have seen in my textbook, and I understand.

$$\tan^{-1}(x)+\tan^{-1}(y)=\pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)\\xy>1$$ This one is mentioned in the book, but not proved.

I can understand the logic behind it, since

  • if the angles related to $x$ and $y$ are such that the angle related to $x+y$ will be in the second quadrant, then it is not possible for the $\tan^{-1}$ to ever give a correct output
  • For $x>0,y>0$ if $xy>1$, then $1-xy<0\Rightarrow\frac{x+y}{1-xy}<0$, and that will give an answer in the fourth quadrant, whereas we likely want one in the second quadrant.

However, this is just intuition. Some questions I have:

  • where does the $xy>1$ come in? How is it established/proved?
  • where does the $\pi$ factor come in?
  • is it possible to see a complete proof for both cases of the property.
K.defaoite
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Starlight
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    The identity in your book only holds for when $x > 0, y > 0$: else $\arctan x + \arctan y = \arctan(\frac{x+y}{1-xy}) - \pi$ instead where $x < 0, y < 0$ (the two cases where $xy > 1$). – Toby Mak Sep 01 '22 at 11:24
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    If you rotate a line through an angle $\theta$ counterclockwise from the $x$ axis, $\tan(\theta)$ gives you the slope of the line. But of course you can always rotate the line another $\pi$ radians (or reverse the rotation by $\pi$ radians) and you will get a line with the same slope. The inverse tangent function works backward from the slope of the line to the rotation angle, but it doesn't "know" how many extra rotations of $\pi$ the line received. That's why you might need to add or subtract $\pi.$ – David K Sep 01 '22 at 12:49
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    https://math.stackexchange.com/questions/1837410/inverse-trigonometric-function-identity-doubt-tan-1x-tan-1y-pi-tan – lab bhattacharjee Sep 01 '22 at 13:38
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    @labbhattacharjee Amazing. This is exactly what I was looking for. Thanks so much. – Starlight Sep 01 '22 at 14:50

2 Answers2

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The identity in your book only holds for when $x > 0, y > 0$: else $\arctan x + \arctan y =$ $\arctan \left(\frac{x+y}{1-xy} \right) - \pi$ instead where $x < 0$ and $y < 0$ (the two cases where $xy > 1$).

The reason is in the sign of $\frac{x+y}{1-xy}$. If $\frac{x+y}{1-xy} > 0$, we either have $x + y > 0$ and $1 - xy > 0$, which implies $y > -x$ and $1 > xy$. In other words, this is the region in between the two branches of $xy = 1$ above the line $y = -x$:

and the region where $\frac{x + y}{1 - xy} < 0$ is the complementary region, not including the line $y = -x$ as that is where $x + y = 0 \implies \frac{x + y}{1 - xy} = 0$.

So when $xy > 1$, $\frac{x+y}{1-xy}$ is negative, and so $\arctan \frac{x+y}{1-xy}$ is also negative, where the domain of $\arctan x$ is the real numbers. But when $x > 0$ and $y > 0$, $\arctan x + \arctan y$ is positive while $\arctan \frac{x+y}{1-xy}$ is negative!

$\arctan x + \arctan y$ will also be greater than $\arctan x + \arctan(1/x) = \pi/2$, and this value is attained (as the infinium) considering that $\arctan 1 + \arctan 1 = \pi/2$. This is because we can think of the choice of $x, y$ as a set of level curves $xy = k$ where $k$ is real: $xy = 1.1, xy = 1.2$ etc. It remains to show that $\arctan x$ is monotonically increasing, so $\arctan(1.1/x) > \arctan(1/x)$ etc. To show this, consider $\tan x$ where $-\frac{\pi}{2} < x < \frac{\pi}{2}$ and reflect it over the line $y = x$.

Hence the right branch of $\arctan x$ to use is the one with range $f(x) \in (\pi/2, 3 \pi/2)$, which is the inverse function of $\tan x$ when $\frac{\pi}{2} < x < \frac{3 \pi}{2}$:

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Compared to the branch of $\tan x$ where $\arctan x$ is usually defined: $f(x) \in (-\pi/2, \pi/2)$, this is a shift of $+\pi$. Hence the formula for $x > 0$ and $y > 0$ has a $+ \pi$ added to it.

Toby Mak
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  • I get the logic behind your post. And this makes it much clearer. But can you explain why the shift is precisely $\pi$ – Starlight Sep 01 '22 at 11:54
  • It comes from considering the minimum value of $\arctan x + \arctan y$ for when $x > 0$ and $y > 0$. This is $\pi/2$ as I have shown, and so of all the different inverses of $\tan x$ with range $(-3 \pi/2, -\pi/2); (-\pi/2, \pi/2); (\pi/2, 3 \pi/2)$ etc., the only one with a minimum of $\pi/2$ is the branch with range $(\pi/2, 3 \pi/2)$, or $\arctan x + \pi$. – Toby Mak Sep 01 '22 at 11:56
  • There is also the same question here, found using Approach0. – Toby Mak Sep 01 '22 at 12:00
  • A simpler way to go about this would be to take an arbitrary choice of $x$ and $y$ such as $x = 2, y = 3$ and check which of the ranges it is in. However, using the fact that $\arctan x + \text{arccot } x$ is just so nice to include. – Toby Mak Sep 01 '22 at 12:11
  • And I'll just mention this last bit: for the other case where $x < 0, y < 0$ and $xy > 1$, the proof is pretty much the same: 1) $\arctan x + \arctan y$ is negative but $\arctan((x+y)/(1 - xy))$ is positive, 2) using the proof here, the maximum value of $\arctan x + \arctan y$ is $-\pi/2$ (reflecting $x, y$ and the inequality region across the $x$-axis), and 3) $\arctan x - \pi$ is the only branch of the inverse that has a maximum value (supremum) of $-\pi/2$. – Toby Mak Sep 01 '22 at 12:20
  • So for the mixed case where $xy < 1$, $\arctan x + \arctan y$ can be positive or it can be negative: $\arctan(-x) = -\arctan x$ after all so you would know that $\arctan 2 + \arctan(-3)$ would be negative based on the monotonicity of $\tan x$. The only branch of the inverse to be both negative and positive is of course, $\arctan x$. – Toby Mak Sep 01 '22 at 12:22
  • I am actually interested in a proof, not a working rule that will get me the answer. – Starlight Sep 02 '22 at 01:38
  • Why do you think this is not a proof? – Toby Mak Sep 02 '22 at 11:04
  • I think your main answer is a proof. I was referring to your comment about the simpler way to go about it. – Starlight Sep 03 '22 at 02:53
  • Ahhhhhh that makes so much more sense now. Thanks for clarifying! – Toby Mak Sep 03 '22 at 03:57
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Here is a proof for $xy>1 $:

Note that the identity $\tan(\tan^{-1} x+\tan^{-1} y)=\frac{x+y}{1-xy}$ leads to $$\tan^{-1} x+\tan^{-1} y= k\pi + \tan^{-1}\frac{x+y}{1-xy} $$ In the case of $x,y>0$, $xy>1\implies y>\frac1{x}$

$$\pi > \tan^{-1} x+\tan^{-1} y > \tan^{-1} x+\tan^{-1} \frac1{x} = \frac\pi2$$

Thus

$$\tan^{-1} x+\tan^{-1} y=\pi+\tan^{-1}\frac{x+y}{1-xy}$$

Quanto
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  • I am not sure I follow along. Which identity are you starting with? And what steps are you taking? in particular where does the $k\pi$ come from? – Starlight Sep 02 '22 at 01:41