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I have to prove $$\arctan(x) + \arctan(y) = \arctan\Bigl(\frac{x+y}{1-xy}\Bigr)$$

Can somebody help me? I don't want you to give me the complete proof, but some start-help would be nice

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    $\begingroup$ Do you know the formulae for $\sin(x+y), \cos(x+y)$ and $\tan (x+y)$? $\endgroup$ – Thomas Nov 15 '14 at 10:03
  • $\begingroup$ yes, but I have no idea to transform $arctan$ in that form $\endgroup$ – Christian Nov 15 '14 at 10:04
  • $\begingroup$ @Christian, This works only if $xy<1$. See math.stackexchange.com/questions/326334/… $\endgroup$ – lab bhattacharjee Nov 16 '14 at 17:27
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From the formula $\tan(\alpha + \beta) = \frac{tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}$, if we let $x = \tan(\alpha)$ and $y = \tan(\beta)$, then

$$ \arctan(x) + \arctan(y) = \alpha + \beta = \arctan(\frac{tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}) = \arctan(\frac{x + y}{1 - xy})$$

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  • $\begingroup$ haha, thank you. I should've known this, the previous proof I had to do was the one you used $\endgroup$ – Christian Nov 15 '14 at 10:06
  • $\begingroup$ You're welcome. :) $\endgroup$ – Empiricist Nov 15 '14 at 10:06
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As the function $\arctan$ satisfies $\tan\arctan x=x$, you just need to prove that $$ \tan(\arctan x+\arctan y)=\tan\arctan\frac{x+y}{1-xy} $$ Just expand the left hand side with the addition formula for the tangent.

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