# $\arctan{\frac{(x+y)}{(1−xy)}}=\arctan x+\arctan y?$-for this to be true xy has to be less than 1,why? [closed]

$$\arctan{\frac{(x+y)}{(1−xy)}}=\arctan x+\arctan y?$$-for this to be true xy has to be less than 1,why?

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## closed as unclear what you're asking by Shuchang, Chasky, Macavity, user1337, Daniel Robert-NicoudDec 1 '13 at 11:58

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what is the question? It is very confusing what you wrote. It does not look like a question to me. Would you mind improving it ? and please, use latex! –  Chasky Dec 1 '13 at 8:58

What is true is that $\tan(\arctan x + \arctan y) = \dfrac{x+y}{1-xy}$ (assuming $xy \ne 1$). So $\arctan x + \arctan y = \arctan \left( \dfrac{x+y}{1-xy} \right) + \pi n$ for some integer $n$. For $(x,y) = (0,0)$ we have $n = 0$, and $\arctan\left( \dfrac{x+y}{1-xy} \right)$ is a continuous function on $\{(x,y): xy < 1\}$, so your equation is true in that region. On the other hand, try points in the two regions of the plane where $xy > 1$ and you'll see $n = +1$ or $n = -1$.
There is no discontinuity in $\arctan((x+y)/(1-xy))$ at $xy = -1$. –  Robert Israel Dec 1 '13 at 22:11