Question

已知tanα=2 ,tanβ= -1/3 ,其中0<α<π/2,π/2<β<π.

1）求tan(α-β)
2）求α+β的值

Answer 1

解(1)：tan(α-β)
=(tanα-tanβ)/(1+tanαtanβ)
=(2+1/3)/(1-2×1/3)
=(7/3)/(1/3)
=7
(2)：tan(α+β)
=(tanα+tanβ)/(1-tanαtanβ)
=(2-1/3)/(1+2×1/3)
=(5/3)/(5/3)
=1
因为0＜α＜π/2，π/2＜β＜π
所以0+π/2＜α+β＜π/2+π
π/2＜α+β＜3π/2
所以tan(α+β)=tan(5π/4)=1
α+β=5π/4