If tan(A-B)tanA+sin2Csin2A=1, A,B,C∈(0,π2), then [2026]
(1)
tanA-tanB(1+tanAtanB)tanA+1+cot2A1+cot2C=1
Put tanA=x, tanB=y, tanC=z
∴ x-y(1+xy)x+(x2+1)z2x2(z2+1)=1
∴ x(x-y)(z2+1)+z2(1+x2)(1+xy)=(1+xy)x2(1+z2)
after solving we get
z2=xy ∵ 1+x2≠0
∴ tan2C=tanA·tanB
∴ tanA, tanC, tanB are in G.P.