已知tan(α+β+
|
∵α+
=α+β+π 3
-(β-π 6
),π 6
∴tan(α+
)=π 3
=tan(α+β+
)-tan(β-π 6
)π 6 1+tan(α+β+
)tan(β-π 6
)π 6
=1
+1 2 1 3 1- 1 6
故答案为1
已知tan(α+β+
|
∵α+
=α+β+π 3
-(β-π 6
),π 6
∴tan(α+
)=π 3
=tan(α+β+
)-tan(β-π 6
)π 6 1+tan(α+β+
)tan(β-π 6
)π 6
=1
+1 2 1 3 1- 1 6
故答案为1