# Proof of an addition theorem

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

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

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