若tan a=2, 则 sina+cosa/sina-cosa+cos^a=?
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![若tan a=2, 则 sina+cosa/sina-cosa+cos^a=?](/uploads/image/z/1596640-40-0.jpg?t=%E8%8B%A5tan+a%3D2%2C+%E5%88%99+sina%2Bcosa%2Fsina-cosa%2Bcos%5Ea%3D%3F)
若tan a=2, 则 sina+cosa/sina-cosa+cos^a=?
若tan a=2, 则 sina+cosa/sina-cosa+cos^a=?
若tan a=2, 则 sina+cosa/sina-cosa+cos^a=?
(sina+cosa)/(sina-cosa)+[cos²a]/[cos²a+sin²a]=[tana+1]/[tana-1]+[1]/[tan²a+1]=16/5