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

How to prove $$\arctan(1)+\arctan(2)+ \arctan(3)=\pi$$
 A: 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!
A: 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
$$\arctan 1+\arctan 2+\arctan 3 = \operatorname {Arg} ((1+i)(1+2i)(1+3i)) = \operatorname {Arg} (-10) = \pi$$
A: 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 $0\cdot\pi=0$
