(1)由sinθ+cosθ=-
,(0<θ<π) 可得,θ 为钝角,且|sinθ|<|cosθ|,故-1<tanθ<0.
10
5
把条件平方可得 sinθcosθ=-
,∴3 10
=-sinθcosθ
sin2θ+cos2θ
,∴3 10
=-tanθ
tan2θ+1
,3 10
即得 tanθ=-
.1 3
(2)(sinθ-cosθ)2=1-2sinθcosθ=
,再由sinθ-cosθ>0,可得 sinθ-cosθ=8 5
=
8 5
.2
10
5
(3)
+1 sinθ
=1 cosθ
=sinθ+cosθ sinθcosθ
=?
10
5 ?
3 10
.2
10
3