Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
Remember that : $$\tan(a+b)=\frac{\tan a + \tan b}{1-\tan a \tan b}.$$ –  user37238 Sep 23 '13 at 7:35
1  
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

2 Answers 2

up vote 6 down vote accepted

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.

share|improve this answer

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 $\cos(a\pm b)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.