已知在⊿ABC中,A,B,C所对的边分别为为a,b,c,若(tanA*tanB)/(tanA+tanB)=1005tanC,且b^2=mc^2,求m的值.
解析：在⊿ABC中,∵(tanA*tanB)/(tanA+tanB)=1005tanC
∴sinAsinB/[sinAcosB+cosAsinB]=1005sinC/cosC
∴sinAsinBcosC=1005sinCsin(A+B)=1005sin^2C
由正弦定理得abcosC=1005c^2,c^2=abcosC/1005．
又∵a^2+b^2=mc^2,
∴a^2+b^2=mabcosC/1005=mab*[(a^2+b^2-c^2)/2ab]/1005=m(a^2+b^2-c^2)/2010
∴m=2010(a^2+b^2)/(a^2+b^2-c^2)= 2010(a^2+b^2)/[a^2+b^2-(a^2+b^2)/m]
∴m(a^2+b^2)- (a^2+b^2)=2010(a^2+b^2)
∴m=2011
已知在⊿ABC中,A,B,C所对的边分别为为a,b,c,且acosB-bcosA=1/2c,当tan(A-B)取得最大值时,求∠C的值.
解析：∵acosB-bcosA=1/2c
由正弦定理化简：
sinAcosB-sinBcosA=1/2sinC=1/2sin(A+B)=1/2(sinAcosB+cosAsinB),
整理得：sinAcosB=3cosAsinB,
两边除以cosAcosB得：tanA=3tanB,
tan(A-B)=(tanA-tanB)/(1+tanAtanB)=2tanB/(1+3tan^2B)=2/[3tanB+1/tanB],
∵A、B是三角形内角,且tanA与tanB同号,
∴A、B都是锐角,即tanA＞0,tanB＞0,
∴3tanB+1/tanB>=2√3,当且仅当3tanB=1/tanB,即tanB=√3/3时取等号,
∴tanA=3tanB=√3,
∴A=π/3,B=π/6,
则C=π/2．
若tan(2x-y)=3tan(x-2y)=3,求tan(x+y)
解析：∵tan(2x-y)=3tan(x-2y)=3
tan(2x-y)=3
tan(x-2y)=1
则tan(x+y)=tan[(2x-y)-(x-2y)]=[ tan(2x-y)-tan(x-2y)]/[1+tan(2x-y)tan(x-2y)]
=(3-1)/(1+3)=1/2