已知sinθ+cosθ=－（根号10）/5，求tanθ的值

sinθ+cosθ=－（根号10）/5
两边平方
(sinθ)^2+2sinθcosθ+(cosθ)^2=2/5
1+2sinθcosθ=2/5
sinθcosθ=-3/10

[(sinθ)^2+(cosθ)^2]/(sinθcosθ)=1/(sinθcosθ)=-10/3

[(sinθ)^2+(cosθ)^2]/(sinθcosθ)
=(sinθ)^2/(sinθcosθ)+(cosθ)^2/(sinθcosθ)
=sinθ/cosθ+cosθ/sinθ
=tanθ+1/tanθ=-10/3
3(tanθ)^2+10tanθ+3=0
(3tanθ+1)(tanθ+3)=0
tanθ=-1/3或tanθ=-3

(sinθ+cosθ)/2^(1/2) = -1/5^(1/2)

sin(θ+PI/4) = -1/5^(1/2)

2kPI + PI < θ+PI/4 < 2kPI + 2PI,

2kPI + 3PI/4 < θ < 2kPI + PI + 3PI/4

当2kPI + 3PI/4 < θ < 2kPI + PI/2 + 3PI/4时，
2kPI + PI < θ+PI/4 < 2kPI + 3PI/2,

cos(θ+PI/4) = -2/5^(1/2)

tan(θ+PI/4) = 1/2

tan(θ) = tan(θ+PI/4 - PI/4)

= [tan(θ+PI/4) - tan(PI/4)]/[1+tan(θ+PI/4)tan(PI/4)]

= [1/2 - 1]/[1 + 1/2]

= -1/3

当2kPI + PI/2 + 3PI/4 <= θ < 2kPI + PI + 3PI/4时，
2kPI + 3PI/2 <= θ+PI/4 < 2kPI + 2PI,