Prove that -
tan(A+B+C) =
tan A+tan B+tan C -tan Atan B tanC/
1-tan Atan B-tan B tan C -tan Ctan A
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Step-by-step explanation:
tan(A+B+C)=sin(A+B+C)/cos(A+B+C)
=sinAcisBcosC+coAsinBcosC+cosAcosBsinC-sinAsinBsinC/cosAcosBcosC-cosAsinBsinC-sinAcosBsinC-sinAsinBcosC
devide cosAcosBcosC in both numerator & denominator
=tanA +tanB+tanC-tanAtanBtanC/1-tanBtanC-tanCtanA-tanAtanB
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