How do you find the exact value of $arc tan(1/2) + arc tan (1/5) + arc tan (1/8)$ ?

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Answer 1 · 2015-06-18T14:48:27

$pi/4rad$

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

Let x= $1/2$ , $y=1/5$ , $z=1/8$

$arctan(1/2)+arctan(1/5)+arctan(1/8)=$

$=arctan$ $((1/2)+(1/5)+(1/8)-(1/2*1/5*1/8))/(1-(1/2*1/5)(1/5*1/8)(1/8*1/2)$

$=arctan(0.8125/0.8125)$

$=arctan(cancel(0.8125)/cancel(0.8125))$

$=arctan(1)$

$=arctan(tan(pi/4))$ [from angle table]

$=cancel(arctan)cancel(tan)((pi/4))$

$=pi/4$

How do you find the exact value of arc tan(1/2) + arc tan (1/5) + arc tan (1/8)arc tan(1/2) + arc tan (1/5) + arc tan (1/8)?