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 |
故答案為5.
8 |
3 |
1 |
3 |
tanA+tanB |
1-tanAtanB |
-
| ||
|
9sin2C-3sinCcosC-5cos2C |
sin2C+cos2C |
9tan2C-3tanC-5 |
tan2C+1 |