证明tana+tanb+tanc

桂子界3270在三角形ABC中,求证tanA+tanB+tanC=tanAtanBtanC -
轩凭庾18172338815 ______[答案] 因为三角形ABC为锐角 所以tanC=tan[ ∏-(A+B)] 即tanC=-(tanA+tanB)÷(1-tanA*tanB) -tanC=(tanA+tanB)÷(1-tanA*tanB) -tanC+tanA*tanB*tanC=tanA+tanB 移项tanA*tanB*tanC=tanA+tanB+tanC

桂子界3270正余弦定理的应用在锐角三角形中证明:tanA+tanB+tanC=tanA*tanB*tanC -
轩凭庾18172338815 ______[答案] ∵tan(A+B)=tanA+tanB/1-tanA*tanB tan(A+B)=tan(π-C)=-tanC ∴tanA+tanB/1-tanA*tanB=-tanC 整理移项即得 tanA+tanB+tanC=tanA*tanB*tanC

桂子界3270有一道高一数学题,请大家帮忙想想在斜三角形ABC中,证明tanA+tanB+tanC=tanAtanBtanC具体过程,不知为啥我算的tanA+tanB+tanC=2tanA+... -
轩凭庾18172338815 ______[答案] 所以tanC=tan[ ∏-(A+B)] 即tanC=-(tanA+tanB)÷(1-tanA*tanB) -tanC=(tanA+tanB)÷(1-tanA*tanB) -tanC+tanA*tanB*tanC=tanA+tanB移项tanA*tanB*tanC=tanA+tanB+tanC...

桂子界3270若A+B+C=180°,证明tanA+tanB+tanC=tanAtanBtanC. -
轩凭庾18172338815 ______[答案] C = 180° - A - B 所以tanC = tan(180° - A - B)= -tan(A + B) 左边 = tanA + tanB - tan(A + B) = tanA + tanB - (tanA + tanB)/(1 - tanAtanB) = [(tanA + tanB)(1 - tanAtanB)- (tanA + tanB)]/(1 - tanAtanB) = -(tanA + tanB)(tanAtanB)/(1 - tanAtanB) = tanAtanB ...

桂子界3270在非直角三角形AB中,求证:tanA+tanB+tanC=tanAtanBtanC -
轩凭庾18172338815 ______[答案] tanC=-tan(A+B) =-(tanA+tanB)/(1-tanAtanB) tanC-tanAtanBtanC=-tanA-tanB tanA+tanB+tanC=tanAtanBtanC 证毕! 此题如仍有疑问,欢迎追问!

桂子界3270一条数学的正切证明题△ABC不是直角三角形求证:tanA+tanB+tanC=tanA·tanB·tanC -
轩凭庾18172338815 ______[答案] 把tan 换做sin/cos,再用正弦定理,余弦定理代入就能反证了.sinA/cosA+sinB/cosB+sinC/cosC=sinAsinBsinC/cosAcosBcosC正弦定理得:a/cosA+b/cosB+c/cosC=abc/cosAcosBcosC通分,然后用余弦定理代入.cosA=b^2+c^2-a^2/2...

桂子界3270已知△ABC是非直角三角形 (1)求证:tanA+tanB+tanC=tanAtanBtanC (已知△ABC是非直角三角形(1)求证:tanA+tanB+tanC=tanAtanBtanC(2)若A>... -
轩凭庾18172338815 ______[答案] 1、三角形ABC tanAtanBtanC=tanA+tanB+tanC证明如下tanA=tan(∏-B-C)=-tan(B+C)= -(tanB+tanC)/(1-tanBtanC)=(tanB+tanC)/(tanBtanC-1)所以 tanA*(tanBtanC-1)=tanB+tanCtanA*tanB*tanC - tanA=tanB+tanC所以tanAta...

桂子界3270证明:tanA+tanB+tanC=tanA*tanB*tanC -
轩凭庾18172338815 ______[答案] 应该是在斜三角形ABC中,求证:tanA+tanB+tanC=tanA*tanB*tanC. ∵tan(A+B)=[tanA+tanB]/[1-tanA*tanB] 且,tan(A+B)=tan(π-C)=-tanC ∴ tanA+tanB/[1-tanA*tanB]=-tanC 整理,得 tanA+tanB=tanC-tanA*tanB*tanC 移项,得 tanA+tanB+tanC=...

桂子界3270求证在三角形ABC中tanA+tanB+tanC=tanAtanBtanC -
轩凭庾18172338815 ______[答案] A+B+C=π 所以tanA=tan[π-(B+C)]=-tan(B+C)=(tanB+tanC)/(tanBtanC-1) 整理得tanA+tanB+tanC=tanAtanBtanC

桂子界3270已知三角形是非直角三角形,abc是三个角,证三角正切的和等于正切的积 -
轩凭庾18172338815 ______[答案] 证明:tan(A+B)=tan(π-C)=-tanC (tanA+tanB)/(1-tanAtanB)=-tanC tanA+tanB=-tanC+tanAtanBtanC 即有,tanA+tanB+tanC=tanAtanBtanC 得证