已知三角形ABC外接圆半径R,且2R(sin^2A-sin^2C)=(√2a-b)sinB成立,求tan(A+B)

题目

已知三角形ABC外接圆半径R,且2R(sin^2A-sin^2C)=(√2a-b)sinB成立,求tan(A+B)

答案

2R(sin^2A-sin^2C)=√2×2RsinAsinB-2RsinBsinB
sinAsinA-sinCsinC=√2×sinAsinB-sinBsinB
sinAsinA-sin(A+B)^2=√2×sinAsinB-sinBsinB
sinAsinA-sinAsinAcosBcosB-sinBsinBcosAcosA-2sinAcosAsinBcosB=√2×sinAsinB-sinBsinB
sinAsinA(1-cosBcosB)-sinBsinBcosAcosA-2sinAcosAsinBcosB=√2×sinAsinB-sinBsiinB
sinAsinAsinBsinB+sinBsinB(1-cosAcosA)-2sinAcosAsinBcosB=√2×sinAsinB
2sinAsinB(sinAsinB-cosAcosB-√2/2)=0
2sinAsinB[-cos(A+B)-√2/2]=0
因为sinA不等于0,sinB不等于0,
所以A+B=135º
所以tan(A+B)=tan135º=-tan45º=1