Is it true that $$ \tan^{-1}\left(\frac{x}{y}\right)+\tan^{-1}\left(\frac{y}{x}\right)=\text{sgn}(xy)\frac{\pi}{2} $$ for $x,y\in\mathbb{R}\backslash \{0\}$? If so, I do I show it? Any hint would be useful.
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What is your motivation? What is the context? Why do you think it is true? – Sahaj Feb 19 '24 at 10:37
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@Sahaj I have simply tried it with multiple values. I am sure there is a straightforward way to show it. I will think about it. This might hint at it. I will edit the post if I get somewhere, no need to close this question. – sam wolfe Feb 19 '24 at 10:38
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You need to show your work and effort. – user Feb 19 '24 at 10:57
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From the post you linked, we know that $$\arctan(t)+\arctan\left(\frac{1}{t}\right)=\frac{\pi}{2}\text{sgn}(t),\quad\forall t\in\mathbb{R}\setminus \{0\}$$ Substituting $t=\frac{x}{y}$ with $y\neq 0$, we get that $$\arctan\left(\frac{x}{y}\right)+\arctan\left(\frac{y}{x}\right)=\frac{\pi}{2}\text{sgn}\left(\frac{x}{y}\right),\quad\forall x,y\in\mathbb{R}\setminus \{0\}$$ Since $\text{sgn}\left(\frac{x}{y}\right)=\text{sgn}\left(xy\right)$, $$\arctan\left(\frac{x}{y}\right)+\arctan\left(\frac{y}{x}\right)=\frac{\pi}{2}\text{sgn}\left(xy\right),\quad\forall x,y\in\mathbb{R}\setminus \{0\}$$
Davide
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