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$$\arctan{\frac{(x+y)}{(1−xy)}}=\arctan x+\arctan y?$$-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-Nicoud Dec 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

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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 $n = -1$.

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considering the range of arctanx+arctany,i got xy<1 but I also got xy>-1,why is the second condition not applicable for the formula to be true? –  PRATIK KUMBHARE Dec 1 '13 at 9:40
There is no discontinuity in $\arctan((x+y)/(1-xy))$ at $xy = -1$. –  Robert Israel Dec 1 '13 at 22:11

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