# 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.

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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

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.

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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 $\cos(a\pm b)$.

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