a-maths!!!!!!!
回答 (1)
2008-01-29 4:42 am
✔ 最佳答案
sin2 θ / 1+ 2cos2 θ = 3/19sin2 θ = (1+ 2cos2 θ) (3/19)
sin2θ = [(1+ 2(1﹣sin2 θ)] (3/19)
19sin2θ= [(9﹣6sin2 θ)]
sin θ = -3/5
cosθ = -4/5
所以，
(tanθ /(1+sec θ )
= (sin θ/cos θ)/[(cos θ+1)/cos θ]
= sin θ/(cos θ+1)
= -3/5/[-4/5+1]
= -3
2008-01-29 17:04:28 補充：
所以，sinθ/(1+cosθ)= (-3/5)/(1-8/5)= (-3/5)/(-3/5)= 1
收錄日期: 2021-04-23 20:33:51
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080128000051KK02925