<s>[INST] What is $-\arctan(\frac{s}{c}) + \pi\Theta(c)$? I know that 
$$-\arctan\left(\frac{1}{c}\right)+\pi\Theta(c)=\arctan(c)+\frac{\pi}{2}$$
where $\Theta(x)$ is the Heaviside step function.
I was wondering if it is possible to find a similar expression for 
$$-\arctan\left(\frac{s}{c}\right) + \pi\Theta(c)$$
 [/INST] Consider a right angled triangle:
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The sum of the angles $\measuredangle CAB$ and $\measuredangle ABC$ is $\frac{\pi}{2}$. We know that $\tan(\measuredangle CAB) = \frac{a}{b}$ and $\tan(\measuredangle ABC) = \frac{b}{a}$ so for $a,b>0$ we have
$$\frac{\pi}{2} = \arctan\left(\frac{b}{a}\right) + \arctan\left(\frac{a}{b}\right)$$
One can extend this to $\frac{a}{b} < 0$ by using that $\tan$ is an odd function so $\arctan(-x) = -\arctan(x)$ giving us
$$\arctan\left(\frac{b}{a}\right) + \arctan\left(\frac{a}{b}\right) = \left\{\matrix{\frac{\pi}{2}, & ab > 0\\ -\frac{\pi}{2} & ab < 0}\right. = -\frac{\pi}{2} + \pi\Theta\left(ab\right)$$
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