可得tanA+tanB=-
| 8 |
| 3 |
| 1 |
| 3 |
所以tan(A+B)=
| tanA+tanB |
| 1-tanAtanB |
-
| ||
|
所以4sin2C-3sinCcosC-5cos2C=9sin2C-3sinCcosC-5cos2C=
| 9sin2C-3sinCcosC-5cos2C |
| sin2C+cos2C |
| 9tan2C-3tanC-5 |
| tan2C+1 |
故答案为5.
| 8 |
| 3 |
| 1 |
| 3 |
| tanA+tanB |
| 1-tanAtanB |
-
| ||
|
| 9sin2C-3sinCcosC-5cos2C |
| sin2C+cos2C |
| 9tan2C-3tanC-5 |
| tan2C+1 |