Proving :$\arctan(1)+\arctan(2)+ \arctan(3)=\pi$ [duplicate]

Question

Possible Duplicate:
Why does $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$?

How to prove $$\arctan(1)+\arctan(2)+ \arctan(3)=\pi$$

Answer 1

As $$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B},$$ $$\tan (A+B+C)=\frac{\tan(A+B)+\tan C}{1-\tan(A+B)\tan C}=\frac{\frac{\tan A+\tan B}{1-\tan A\tan B}+\tan C}{1-\tan C\left(\frac{\tan A+\tan B}{1-\tan A\tan B}\right)}=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}$$

So, $$A+B+C=\arctan\left(\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}\right)+m\pi$$ where $m$ is any integer.

Putting $\tan A=1, \tan B=2,\tan C=3$,

So, $$\arctan 1+m_1\pi+\arctan 2y+m_2\pi+\arctan 3+m_3\pi=\arctan 0+m\pi$$ where $m_i$ are integers.

The general value of $\arctan 1+\arctan 2+\arctan 3=\arctan 0+(m-m_1-m_2-m_3)\pi$ $=(n+m-m_1-m_2-m_3)\pi=r\pi$ where $n$ is any integer, hence $r=n+m-m_1-m_2-m_3$ is.

As the special value of each of the inverse trigonometric function lies in $\left(0,\frac\pi2\right)$ so their sum will lie in $(0,\frac{3\pi}2).$ Hence, the special value being $0\cdot\pi=0$

Answer 2

Once I have seen a very nice proof of this: your claim is equivalent to proving that the sum of red, green and blue angles is $\pi$. Note that in the second picture, the blue-green triangle is right and isosceles (that is 45-45-90 triangle and thus similar to small red-black triangle in the first diagram).

Cheers!

Answer 3

Consider $(1+i)$, $(1+2i)$ and $(1+3i)$ which have respective arguments of $\arctan 1$, $\arctan 2$ and $\arctan 3$. Their product equals the sum of the arguments. Thus

$$\arctan 1+\arctan 2+\arctan 3 = \operatorname {Arg} ((1+i)(1+2i)(1+3i)) = \operatorname {Arg} (-10) = \pi$$