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Question

$If,A+B+C=npi$
then prove that, $color(brown)(tanA+tanB+tanC=tanAtanBtanC)$

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Answer 1 · 2018-01-12T09:17:57

$Given A+B+C=npi$

$=>A+B=npi-C$

$=>tan(A+B)=tan(npi-C)$

$=>(tanA+tanB)/(1-tanAtanB)=-tanC$

$=>(tanA+tanB)=-(1-tanAtanB)tanC$

$=>(tanA+tanB)=-tanC+tanAtanBtanC$

, $color(brown)(=>tanA+tanB+tanC=tanAtanBtanC)$

Answer 2 · 2018-01-12T09:20:00

Please see below.

As $A+B+C=npi$

$A+B=npi-C$

and $tan(A+B)=tan(npi-C)$

or $(tanA+tanB)/(1-tanAtanB)=-tanC$

or $tanA+tanB=-tanC+tanAtanBtanC$

or $tanA+tanB+tanC=tanAtanBtanC$