已知sinα+cosα=(根号10)/5,求tanα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=－（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)
=(s...