Comments on Show that $f(x) = \arctan\left(\frac{x}{x+1}\right) + \arctan\left(\frac{x+1}{x}\right) = \frac{\pi}{2}$
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Show that $f(x) = \arctan\left(\frac{x}{x+1}\right) + \arctan\left(\frac{x+1}{x}\right) = \frac{\pi}{2}$
Show that $$f(x) = \arctan\left(\frac{x}{x+1}\right) + \arctan\left(\frac{x+1}{x}\right) = \frac{\pi}{2} \quad \forall x \in (-\infty, -1)\large\cup (0, \infty)$$
I observed this feature graphically, but am looking for a way to prove it. My attempt is as follows:
We can show that $$\arctan(A) + \arctan(B) = \arctan\left(\frac{A+B}{1 - AB}\right)$$ It follows that
$$\displaylines{f(x) = \arctan\left(\frac{x}{x+1}\right) + \arctan\left(\frac{x+1}{x}\right) = \arctan\left(\frac{\frac{x}{x+1} + \frac{x+1}{x}}{1 - \left(\frac{x}{x+1}\right)\left(\frac{x+1}{x}\right)}\right) \cr = \arctan\left(\frac{\frac{x}{x+1} + \frac{x+1}{x}}{1 - 1}\right) = \arctan\left(\frac{\frac{x}{x+1} + \frac{x+1}{x}}{0}\right) = \arctan(\infty) = \boxed{\frac{\pi}{2}}}$$
My attempt has a few questionable steps. The first is the deduction that division by 0 yields $\infty$. Secondly, I have been unable to consider the appropriate domain. Although my steps exclude for $-1$ and $0$, I don’t think they accurately consider values in between. I would appreciate if someone can make this "intuitive" proof more rigorous/concrete. I am also open to alternate solutions, but I would also appreciate comments on my attempt.
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You're right that division by zero is a problem if you want to use that identity. Rather than salvage that, I would recommend this identity instead:
$$ \arctan a + \arctan{a^{-1}} = \frac\pi2\qquad\text{for positive $a$} $$
Proving this identity is easy: draw a $1 \times a$ rectangle with a diagonal and consider the two triangles thus formed.
Then show that $\frac{x}{x+1}$ is positive for all $x$ in the domain of $f$ and you're done.
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