How do you find the exact value of $arctan(1) + arctan(2) + arctan(3)$ ?

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Answer 1 · 2015-06-26T03:13:47

Answer is 0.

$arctanx+arctany+arctanz=arctan$ $(x+y+z-xyz)/(1-xy-yz-zx)$

Let x= $1$ , $y=2$ , $z=3$

$arctan(1)+arctan(2)+arctan(3)=$

$=arctan$ $((1)+(2)+(3)-(1*2*3))/(1-(1*2)(2*3)(3*1)$

$=arctan(0/(-35))$

$=arctan(0)$

$=arctan(tan0))$ [from angle table]

$=cancel(arctan)cancel(tan)((0))$

$=0$

How do you find the exact value of arctan(1) + arctan(2) + arctan(3) arctan(1) + arctan(2) + arctan(3) ?