# Correct my proof that: $\tan^{-1}(1) + \tan^{-1}(2) +\tan^{-1}(3) = \pi$ [duplicate]

LHS = $\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3)$

= $\frac{\pi}{4}+ \tan^{-1}\left(\frac{2+3}{1-2.3}\right)$

= $\frac{\pi}{4}+\tan^{-1}{5\over-5}$

= $\frac{\pi}{4}+ \tan^{-1}(-1)$

=$\frac{\pi}{4}-\tan^{-1}(1)$

=$\frac{\pi}{4}-\frac{\pi}{4}$

= 0

Where am i going wrong?

• The equation $y=\tan x$ doesn't have a single solution in $x$...
– user65203
Feb 3, 2017 at 16:28

Note that we have that $$\arctan x +\arctan y =\arctan \frac{x+y}{1-xy} \pmod {\pi}$$ The $\mod \pi$ appears because the tangent function has a period of $\pi$. So it is not necessarily guranteed that $$\arctan x +\arctan y =\arctan \frac{x+y}{1-xy}$$ This is especially true since $\arctan 2$ and $\arctan 3$ are both positive, but $-\frac{\pi}{4}$ is negative in your case. So we should actually have $\arctan 2 +\arctan 3=\frac{3 \pi}{4}$.

Note that $\arctan x=\tan^{-1} x$. I add this so as to avoid confusion in notation.

• "The $\mod \pi$ appears because the tangent function has a period of $\pi$" What is meant by 'function has a period of $\pi$' and $\mod \pi$? Feb 3, 2017 at 16:17
• @mikealise A function has a period in $\pi$ means that the function satisfies $$f(x+\pi)=f(x)$$ Also $x=y \pmod {\pi}$ if $x-y$ is a integer times $\pi$. So, for example $$\pi+1 =1 \pmod {\pi}$$ Feb 3, 2017 at 16:20
• @mikealise Do you understand? Feb 3, 2017 at 16:22

Not actually showing what you did wrong. S.C.B. did that. Just a nice graphical proof of your identity.

• The graphical proof that can be found here is a little simpler. Mar 13, 2023 at 17:49

$$\tan(\arctan1+\arctan2+\arctan3)=\frac{1+\tan(\arctan2+\arctan3)}{1-\tan(\arctan2+\arctan3)}=\frac{1+\dfrac{2+3}{1-2\cdot3}}{1-\dfrac{2+3}{1-2\cdot3}}=\frac{1-1}{1+1}=0.$$

This implies

$$\arctan1+\arctan2+\arctan3=k\pi$$ for some $k$.

As the arc tangents are in range $(0,\pi/2)$, we have

$$0<k\pi<\frac{3\pi}{2}$$ which is conclusive.