第一问考查三角形恒等式tanA+tanB+tanC=tanAtanBtanC
显然tanC=-1
2)由正弦定理得a/sinA=b/sinB=2R=2
a=2sinA,b=2sinB,A+B=π/4,
S△ABC=absinC/2=√2sinAsinB=√2sinAsin(π/4-A)
=sinA(cosA-sinA)=sinAcosA-(sinA)^2=sin2A/2+cos2A-1/2
=√2sin(2A+π/4)-1/2,
A∈(0,π/4)则2A+π/4∈(π/4,3π/4),
√2/2<sin(2A+π/4)<1,
∴0<S<√2-1/2,
Smax=√2-1/2
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