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How to show that for all $(x,y)\in\mathbb R^2\setminus (-\infty, 0]\times \{0\}$ $$ \arctan\left(\dfrac{x^2+y^2-x}{y}\right)-\arctan\left(\dfrac{x-1}{y}\right) = 2\arctan\left(\dfrac{y}{x+\sqrt{x^2+y^2}}\right) $$ I don't see how to proceed... :/

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    A fun trick: Show that the derivative with respect to $x$ is the same on both sides. – 2017-02-27
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    Do you have your domain right? The left hand side is undefined on $(0,\infty) \times {0}$. – 2017-02-27
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    For me, it's defined because I have this identity $\int_0^1 \dfrac{y}{(t(x-1)+1)^2 + t^2y^2}\,\mathbf dt = \arctan\left(\dfrac{x^2+y^2-x}{y}\right)-\arctan\left(\dfrac{x-1}{y}\right)$... ? – 2017-02-27
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    In single variable calculus if $f'(x) = g'(x)$ and f(x) and g(x) agree at 1 point, then the two functions are identically equal by the fundamental theorem of calculus. A similar statement should be true for functions of two variables. – 2017-02-27
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    If you plug in (1,0) into the right hand side of the equation in your question and if you plug in (1,0) into the integral in your comment then you also get 0. – 2017-02-27
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    I'm not sure if this approach works, but it is an idea. – 2017-02-27
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    Thank you @Simply Beautiful Art, I've already seen this method to prove that for $x>0, \arctan x +\arctan(1/x)=\pi/2$ :) – 2017-02-27
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    @Phoenix Aw dang! I was about to mention that! >.> – 2017-02-27

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