<s>[INST] 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 $$\arctan(1)+\arctan(2)+ \arctan(3)=\pi$$
 [/INST] 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 $0\cdot\pi=0$
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