Is my trig result unique? 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!

If you differentiate the function $$f(t)=\arctan t - \arctan\frac{1 + t}{1 - t},$$ you get zero, so the function is constant in each of the two intervals $(-\infty,1)$ and $(1,+\infty)$ on which it is defined. 


*

*Its value at zero is $\pi/2$, so that $f(t)=-\pi/4$ for all $t<1$, so
$$ \arctan t + \frac\pi4 = \arctan\frac{1 + t}{1 - t},\qquad\forall t<1.$$

*On the other hand, one easily shows that $\lim_{t\to+\infty}f(t)=\frac{3\pi}{4}$, so 
$$ \arctan t - \frac{3\pi}4 = \arctan\frac{1 + t}{1 - t},\qquad\forall t>1.$$
If $t=a/b$ is a rational number smaller that $1$, then the first point is your identity. If it larger than $1$, we see that you have to change things a bit.
