已知sinα-cosα=-根号5/5,求tanα的值

由象限范围可知sinα和cosα同号,则tanα大于0.不妨设α在第一象限,则sinα大于0,cosα大于0.
2sinαcosα=(sinα^2+cosα^2)-(sinα-cosα)^2=1-(sinα-cosα)^2
(sinα+cosα)^2=(sinα-cosα)^2+4sinαcosα=2-(sinα-cosα)^2=9/5
sinα+cosα=3√5/5
sinα-cosα=-√5/5
解得sinα==√5/5 cosα==2√5/5
tanα=1/2