问题
解答题
已知tan(α+
|
答案
tan(β+
)=tan[(a+π 3
)-(α-β)]=π 3
=tan[(α+
)-tan(α-β)]π 3 1+tan(α+
)•tan(α-β)π 3
=
-1 3 1 4 1+
•1 3 1 4
.1 13
已知tan(α+
|
tan(β+
)=tan[(a+π 3
)-(α-β)]=π 3
=tan[(α+
)-tan(α-β)]π 3 1+tan(α+
)•tan(α-β)π 3
=
-1 3 1 4 1+
•1 3 1 4
.1 13