Given △ ABC, (Tana + 1) (tanb + 1) = 2, ab = 2, find: (1) the degree of angle c; (2) the maximum area of triangle ABC

The opposite sides of corner a, corner B and corner C are a, B and C (1) Tana + Tana + tanb + tanataanb + 1 = 2, that is, Tana + tanb = 1-tanataanb, \\\cornera, corner B and corner C are a, B and C (1) Tana + Tana + tanb + tanb + tanataanb + 1 = 2, namely, a, B, C (1) Tana + Tana + Tana + tanb = 1 + Tana + tanb + tanb + tanb + tanb + tanatab + tanataanb + tanataanb + 1 + Tana + tanb + Tana + Tana + B + B + B + B + B (1) (Tanc [π (0, π), (0, π), (0, π), (0, (0, π), (0,π) \\\\a ≥ 2Ab, That is ab ≤ 4-22, so s △ ABC = 12absinc = 24ab ≤ 24 (4-22) = 2-1

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