Question

已知sinx+cosx/sinx-cosx=3,求tanx,2sin^2x+(sinx-cosx)^2的值？

Answer 1

因为sinx+cosx/sinx-cosx=3,
所以
sinx+cosx=3sinx-3cosx
sinx=2cosx
即
tanx=sinx/cosx=2
又sin²x+cos²x=1
5cos²x=1
cos²x=1/5
所以
2sin^2x+(sinx-cosx)^2
=2-2cos²x+1-2sinxcosx
=2-2cos²x+1-4cos²x
=3-6cos²x
=3-6/5
=9/5,5,0,（sinx+cosx）/（sinx-cosx）=3
上下同除以cosx得
(tanx+1)/(tanx-1)=3
解得
tanx=2
2sin^2x+(sinx-cosx)^2
=[2sin^2x+(sinx-cosx)^2]/1
=[2sin^2x+(sinx-cosx)^2]/(sin^2x+cos^2x) (上下同除以cos^2x)
=[2tan^2x+(tanx-1)^2]/(tan^2x+1)
=[8+1]/5
=9/5,0,