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I recently determined that for all integers $a$ and $b$ such that $a\neq b$ and $b\neq 0$,

$$ \arctan\left(\frac{a}{b}\right) + \frac{\pi}{4} = \arctan\left(\frac{b+a}{b-a}\right) $$

This implies that 45 degrees away from any angle with a rational value for tangent lies another angle with a rational value for tangent. The tangent values are related.

If anyone can let me know if this has been done/shown/proven before, please let me know. Thanks!

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    There are other ways to prove that 45 degrees from a rational slope lies a rational slope, and it is not very hard to do, but I must say that I have never seen this identity before. Welcome to MSE! =) +1.2011-12-13
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    Remember that $\arctan$ (the inverse tangent) always takes values between $-\pi/2$ and $\pi/2$. If you pick a rational that is greater than $\pi/4$ and less than $\pi/2$, then the left hand side of your question cannot be equal to the value of the arctangent at *any* point, let alone at a rational point. So what you write is not what you meant to write. What you mean, I think, is that if $\alpha$ is an angle such that $\tan(\alpha)=\frac{a}{b}$, then $\tan(\alpha+\frac{\pi}{4}) = \frac{b+a}{b-a}$.2011-12-13
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    @Arturo : Perhaps that is what OP did in his proof? Maybe we should ask him how he did this and help him on his definitions so that such details might not slip his mind again. After all he had an idea in mind.2011-12-13
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    +1 for research. Praise, not shame, for the (re)discovery.2016-06-05

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