设tan
| α |
| 2 |
| 2x |
| 1+x2 |
| 1−x2 |
| 1+x2 |
| 2x |
| 1−x2 |
∵α∈[0°,180°],∴
| α |
| 2 |
∴tan
| α |
| 2 |
∵sinα+cosα=
| 1 |
| 5 |
∴
| 2x |
| 1+x2 |
| 1−x2 |
| 1+x2 |
| 1 |
| 5 |
∴6x2-10x-4=0,
解得x=2,或x=-
| 1 |
| 3 |
因此tanα=
| 2x |
| 1−x2 |
| 4 |
| 3 |
∴直线的斜率k=tanα=-
| 4 |
| 3 |
故选C
| 1 |
| 5 |
| 4 |
| 3 |
| 3 |
| 4 |
| 4 |
| 3 |
| 3 |
| 4 |
| 4 |
| 3 |
| α |
| 2 |
| 2x |
| 1+x2 |
| 1−x2 |
| 1+x2 |
| 2x |
| 1−x2 |
| α |
| 2 |
| α |
| 2 |
| 1 |
| 5 |
| 2x |
| 1+x2 |
| 1−x2 |
| 1+x2 |
| 1 |
| 5 |
| 1 |
| 3 |
| 2x |
| 1−x2 |
| 4 |
| 3 |
| 4 |
| 3 |