# Why is $\arctan\frac{x+y}{1-xy} = \arctan x +\arctan y$?

Why is $\arctan\frac{x+y}{1-xy} = \arctan x +\arctan y$?

It is said that this is derived from trigonometry, but I couldn't find why this is the case.

• Remember that : $$\tan(a+b)=\frac{\tan a + \tan b}{1-\tan a \tan b}.$$ – user37238 Sep 23 '13 at 7:35
• This is not quite true. What is true is that $\tan(s+t)=\frac{\tan s+\tan t}{1-\tan s\tan t}$. The reason the $\arctan$ version can fail is because of restrictions on the range of $\arctan$. But it is true, for example, when $0\le x,y\lt \frac{\pi}{4}$. – André Nicolas Sep 23 '13 at 7:38
• – lab bhattacharjee Jun 24 '16 at 4:57
• Because they are the same number :p – Jacob Wakem Aug 6 '16 at 17:00

Here's a proof that has essentially nothing to do with trigonometry:

Hold $y$ constant and differentiate the function $$f(x) = \arctan{\frac{x + y}{1 - xy}} - \arctan{x} - \arctan{y}$$ to find that

\begin{align}f'(x) &= \frac{1}{1 + \left(\frac{x + y}{1 - xy}\right)^2} \cdot \left(\frac{(1 - xy) - (x + y)(-y)}{(1 - xy)^2}\right) - \frac{1}{1 + x^2} \\ &= \frac{(1 - xy)^2}{(1 - xy)^2 + (x + y)^2} \cdot \left(\frac{1 - xy + xy + y^2}{(1 - xy)^2}\right) - \frac{1}{1 + x^2} \\ &= \frac{1 + y^2}{1 - 2xy + x^2y^2 + x^2 + 2xy + y^2} - \frac{1}{1 + x^2} \\ &= \frac{1 + y^2}{1 + x^2 + y^2 + x^2y^2} - \frac{1}{1 + x^2} \\ &= \frac{1 + y^2}{(1 + y^2) + x^2 (1 + y^2)} - \frac{1}{1 + x^2} \\ &= \frac{1}{1 + x^2} - \frac{1}{1 + x^2} = 0 \end{align}

So $f$ is a constant. Letting $x = 0$, we find that $f(0) = 0$ and the identity follows.

• This is indeed true for $\mid xy \mid <1$, thats why you can take $x=0$. But what if we have $\mid xy \mid>1$? Then you should take some x so that it is always true and you should get $f(0)=-\pi$. – HeatTheIce Jan 20 '16 at 16:49

The following identity is useful for your purpose

$$\tan(\alpha \pm \beta) = \frac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta}.$$

To prove it just use the identity $\tan(t) = \frac{\sin(t)}{\cos(t)}$ and the identities for $\sin(a\pm b)$ and $\cos(a\pm b)$.

Computing with complex numbers: $$(1+xi)(1+yi)=(1-xy)+(x+y)i$$ Take arg on both sides $$\arctan x + \arctan y = \arg(1+xi)+\arg(1+yi) = \arg((1+xi)(1+yi)) \\ = \arg((1-xy)+(x+y)i) = \arctan\frac{x+y}{1-xy}$$

• A very illuminating approach. Most trig relationships can be trivially derived using the Euler's identity and cartesian to polar conversion like this. – orion Oct 3 '18 at 12:15

Do NOT blindly use this for calculations. Example $\arctan(5000)+\arctan(5000) \sim \pi/2+\pi/2=\pi$ with the formula quoted you get $\arctan(10000/(1-25\times10^6))\sim\arctan(-1/2500)\sim~0$. Like Andre Nicolas said, you need to considr arctan ranges

Consider $$A=\arctan(x)+\arctan(y),$$ then $$\tan(A)=\tan(\arctan(x)+\arctan(y)),$$ which is also equal to:$$\frac{\tan(\arctan(x))+\tan(\arctan(x))}{1-\tan(\arctan(x)) + \tan(\arctan(y)},$$ which simplifies to giv:$$\tan(A)=\frac{x+y}{1-xy}.$$

Thus,$$\arctan(x)+\arctan(y)=A=\arctan(\tan(A))=\arctan\frac{x+y}{1-xy}.$$

Ok fine, mlc. I'll answer the question. Here is a trigonometric proof. No calculus required. Statement: $\arctan(\frac {x+y} {1-xy})=\arctan(x)+\arctan(y)$ whenever $|xy|<1$.

Proof. Choose $(x,y)\in\big\{|xy|<1\big\}$ arbitrarily. This formula is obvious if $x=0,$ so first consider when $x>0$. We prove that $\arctan(x)+\arctan(y)\in\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)$. Note $|xy|<1$ if and only if $-\frac{1}{x}<y<\frac1x$. Since $\arctan$ is an increasing and odd function, it follows $$-\arctan\Big(\frac{1}{x}\Big)<\arctan(y)<\arctan\Big(\frac{1}{x}\Big).$$ Indeed, $\arctan(x)+\arctan(\frac{1}{x})=\frac{\pi}{2},$ and so $$-\frac\pi2<2\arctan(x)-\frac\pi2<\arctan(x)+\arctan(y)<\frac\pi2.$$ This shows $\arctan(x)+\arctan(y)$ belongs to the range of $\arctan$, $\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)$. The angle sum identity tells us $$$$\begin{split}\tan(\arctan(x)+\arctan(y)) &=\frac{\tan(\arctan(x))+\tan(\arctan(y))}{1-\tan(\arctan(x))\tan(\arctan(y))} \\ & =\frac{x+y}{1-xy}\end{split}$$$$

Hence $$$$\begin{split}\arctan(x)+\arctan(y)&=\arctan(\tan(\arctan(x)+\arctan(y))) \\ &=\arctan\Big(\frac{x+y}{1-xy}\Big)\end{split}$$$$ This proves the identity for $x>0,$ so now assume $x<0$. Then $-x>0$ and $|xy|<1$ $\iff$ $-\frac{1}{x}<-y<\frac{1}{x}.$ Finally, $$$$\begin{split}\arctan(x)+\arctan(y) & =-\big(\arctan(-x)+\arctan(-y)\big) \\ & =-\arctan\Big(\frac{-x-y}{1-(-x)(-y)}\Big) \\ & =\arctan\Big(\frac{x+y}{1-xy}\Big)\end{split}$$$$ The result follows.