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三角形ABC三边ABC成等比数列,CosB=3/4,求1/tAnA+...

解: B为三角形内角,sinB恒>0 cosB=3/4 sinB=√(1-cos²B)=√[1-(3/4)²]=√7/4 a、b、c成等比数列,则b²=ac 由正弦定理得sin²B=sinAsinC 1/tanA +1/tanC =cosA/sinA +cosC/sinC =(sinAcosC+cosAsinC)/(sinAsinC) =sin(A+C)/(si...

a,b,c成等比数列 ac=b^2 By cosine rule cosB= (a^2+c^2-b^2)/(2ac) 3/4= (a^2+c^2-ac)/(2ac) 3ac=2(a^2+c^2-ac) a^2-2ac +c^2=0 a=c cosB=3/4 =>sinB=√7/4 By sine rule a/sinA=c/sinC sinC/sinA = c/a =1 1/tanA+1/tanB =cosA/sinA+ cosB/si...

1. a,b,c成等比数列,ac=b^2,sinA*sinC=sinB^2 (a/sinA=Bb/sinB=c/sinC=2R) cotA+cotC= cosA /sinA +cosC /sinC =(cosA sinC +cosC sinA )/sinA sinC =sin(A+C )/sinB ^2 =sinB /sin B^2 =1/sinB =根号(1-cosB^2)=根号7/4 2. a、b、c成等比数列...

解:(1)1/tanA+1/tanC=(tanA+tanC)/(tanA*tanC)=cosA*cosC(tanA+tanC)/(sinA*sinC) =(sinAcosC+sinCcosA)/(sinA*sinC)=[sin(A+C)]/(sinA*sinC) ∵cosB=3/4 又∵在△ABC中,00 ∴a+c=3

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