tanα=2/3(sin²α-2sinαcosα+1)/(sin²a+cos²a)=(2sin²a-2sinacosa+cos²a)/(sin²a+cos²a)除以cos²a得=(2tan²a-2tana+1)/(tan²a+1)=(8/9-4/3+1)/(4/9+1)=(5/9)/(...