题目内容
(理)若sin(α-β)cosα-cos(α-β)sinα=
,β在第三象限,则tan(β+
)=______.
(文)已知α∈(
,π),sinα=
,则tan(α+
)=______.
2
| ||
| 3 |
| π |
| 4 |
(文)已知α∈(
| π |
| 2 |
| 3 |
| 5 |
| π |
| 4 |
(理)∵sin(α-β)cosα-cos(α-β)sinα=
,
∴sin[(α-β)-α]=
,即sinβ=-
,
又∵β在第三象限,∴cosβ=-
=-
,则tanβ=2
,
∴tan(β+
)=
=-
;
(文)∵α∈(
,π),sinα=
,
∴cosα=-
=-
,则tanα=-
,
∴tan(α+
)=
=
.
故答案为:-
,
.
2
| ||
| 3 |
∴sin[(α-β)-α]=
2
| ||
| 3 |
2
| ||
| 3 |
又∵β在第三象限,∴cosβ=-
1-
|
| 1 |
| 3 |
| 2 |
∴tan(β+
| π |
| 4 |
tanβ+tan
| ||
| 1-tanβ |
9+4
| ||
| 7 |
(文)∵α∈(
| π |
| 2 |
| 3 |
| 5 |
∴cosα=-
1-
|
| 4 |
| 5 |
| 3 |
| 4 |
∴tan(α+
| π |
| 4 |
tanα+tan
| ||
| 1-tanα |
| 1 |
| 7 |
故答案为:-
9+4
| ||
| 7 |
| 1 |
| 7 |
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