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

Solution 1:

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.

Solution 2:

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 $\cos(a\pm b)$.

Solution 3:

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

Solution 4:

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