已知▲ABC是非直角三角形,求证tanA+tanB+tanC=tanAtanBtanC
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已知▲ABC是非直角三角形,求证tanA+tanB+tanC=tanAtanBtanC
tanA+tanB+tanC
=sinA/cosA+sinB/cosB+sinC/cosC
=(sinAcosBcosC+sinBcosAcosC+sinCcosAcosB)/cosAcosBcosC
=(cosC(sinAcosB+sinBcosA)+sinCcosAcosB)/cosAcosBcosC
=(cosCsin(A+B)+sinCcosAcosB)/cosAcosBcosC
=(cosCsinC+sinCcosAcosB)/cosAcosBcosC
=sinC(cos(π-A-B)+cosAcosB)/cosAcosBcosC
=sinC(-cos(A+B)+cosAcosB)/cosAcosBcosC
=sinC(-cosAcosB+sinAsinB+cosAcosB)/cosAcosBcosC
=sinCsinAsinB/cosAcosBcosC
=tanAtanBtanC
=sinA/cosA+sinB/cosB+sinC/cosC
=(sinAcosBcosC+sinBcosAcosC+sinCcosAcosB)/cosAcosBcosC
=(cosC(sinAcosB+sinBcosA)+sinCcosAcosB)/cosAcosBcosC
=(cosCsin(A+B)+sinCcosAcosB)/cosAcosBcosC
=(cosCsinC+sinCcosAcosB)/cosAcosBcosC
=sinC(cos(π-A-B)+cosAcosB)/cosAcosBcosC
=sinC(-cos(A+B)+cosAcosB)/cosAcosBcosC
=sinC(-cosAcosB+sinAsinB+cosAcosB)/cosAcosBcosC
=sinCsinAsinB/cosAcosBcosC
=tanAtanBtanC
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