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A+B+C=pi/2=>tanAtanB+tanBtanC+tanCtanA=...

`A+B+C=pi/2=>tanAtanB+tanBtanC+tanCtanA=`

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Given that tan(A+B)=(tanA+tanB)/(1-tanAtanB) by writing A+B+C=(A+B)+C prove that tan(A+B+C)=(tanA+tanB+tanC-tanAtanBtanC)/(1-(tanAtanB+tanBtanC+tanCtanA)

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