在△ABC中A,B,C满足cosB+sinCcosA=0 (1)用tanA表示tanC (2)求角B范围.

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在△ABC中A,B,C满足cosB+sinCcosA=0 (1)用tanA表示tanC (2)求角B范围.
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在△ABC中A,B,C满足cosB+sinCcosA=0 (1)用tanA表示tanC (2)求角B范围.
在△ABC中A,B,C满足cosB+sinCcosA=0 (1)用tanA表示tanC (2)求角B范围.

在△ABC中A,B,C满足cosB+sinCcosA=0 (1)用tanA表示tanC (2)求角B范围.
(1)将cosB换作cos(180-(A+C))=-cos(A+C)代入展开后同除cosAcosC得tanC=1/(1+tanA) (2)tanB=-tan(A+C)=(tanA+tanC)/(tanAtanC-1).然后将上问所得代入可得tanB=-(tanA)2-tanA-1由二次函数最值得tanB

cosB+sinCcosA=0
cosB=-cos(A+C)=sinAsinC-cosAcosC
sinAsinC-cosAcosC+sinCcosA=0
cosAcosC(tanA-1+tanC)=0
因为cosAcosC不等于0,所以tanA+tanC-1=0
tanC=1-tanA
tan(A+C)=(tanA+tanC)/(1-tanAtan...

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cosB+sinCcosA=0
cosB=-cos(A+C)=sinAsinC-cosAcosC
sinAsinC-cosAcosC+sinCcosA=0
cosAcosC(tanA-1+tanC)=0
因为cosAcosC不等于0,所以tanA+tanC-1=0
tanC=1-tanA
tan(A+C)=(tanA+tanC)/(1-tanAtanC)=1/(1-tanAtanC)
因为00
tanA+tanC=1,tanAtanC<=1/4
故1pi/4180-arctan(4/3)<=B<=3/4*pi

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