Let a, B, C of △ ABC be opposite to a, B, C, and acosb bcosa = 2C. (I) prove that Tana = - 3tanb; (II) find the maximum of angle C

Let a, B, C of △ ABC be opposite to a, B, C, and acosb bcosa = 2C. (I) prove that Tana = - 3tanb; (II) find the maximum of angle C

(I) in △ ABC, from the sine theorem and acosb bcosa = 2c, (2 points), we can get sinacosb sinbcosa = 2sinc = 2Sin (a + b) = 2sinacosb + 2cosasinb ｛ sinacosb = - 3cosasinb, so Tana = - 3tanb; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (4 points) (II) from Tana = - 3tanb, we can see that one of a and B must be an obtuse angle and the other an acute angle; (6 points) suppose B is an obtuse angle, then acosb bcosa = 2C ＜ 0, which is contrary to the known, so B must be an acute angle, a is an obtuse angle, ∵ a + B + C = π, so Tanc = - Tan (a + b) = Tana + tanbtanatanb − 1, Tana = - 3tanb is substituted, and Tanc = 21tanb + 3tanb ≤ 33, (8 points) so C ≤ π 6, if and only if 3tanb = 1tanb, that is, tanb = 33, then the equal sign holds, Tana = - 3, that is, when a = 2 π 3, B = π 6, C gets the maximum Value π 6. & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (12 points)

In △ ABC, the opposite sides of angles a, B and C are a, B and C respectively, if bcosa acosb = 12C. (I) prove that tanb = 3tana; (II) if Tanc = 2, find the value of angle A