化简: (1)sin(α+β)−2sinαcosβ2sinαsinβ+cos(α+β); (2)1/1−tanθ-1/1+tanθ.

问题描述:

化简:
(1)

sin(α+β)−2sinαcosβ
2sinαsinβ+cos(α+β)

(2)
1
1−tanθ
-
1
1+tanθ

(1)原式=

sinα•cosβ+cosα•sinβ−2sinα•cosβ
2sinα•sinβ+cosα•cosβ−sinα•sinβ

=
−(sinα•cosβ−cosα•sinβ)
cosα•cosβ+sinα•sinβ

=-
sin(α−β)
cos(α−β)
=-tan(α-β).
(2)原式=
(1+tanθ)−(1−tanθ)
1−tan2θ

=
2tanθ
1−tan2θ
=tan2θ.