更新1:
唔該，但係只學到sin, cos and tan 呢 3 款，所以可唔可以用呢3款再教我呢?謝謝!
急! 請證明: 1 + tan^2 (90^0+θ) = tan 270^0 +θ / cos (180^ + θ) sin (360^0 +θ)?
回答 (2)
2016-06-12 10:01 am
✔ 最佳答案
1+Tan^2_(90度+θ)=Tan(270度+θ) /[Cos(180度+θ)*Sin (360度+θ)]?
Sol
1+Tan^2_(90度+θ)
=1+Cot^2_θ
=Csc^2_θ
Tan(270度+θ)/[Cos (180度+θ)*Sin (360度+θ)]
=－Cotθ/[(－Cosθ)*Sinθ]
=Cotθ/(Cosθ*Sinθ)
=Sinθ*Cotθ/(Sinθ*Cosθ*Sinθ)
=Cosθ/(Sinθ*Cosθ*Sinθ)
=1/(Sinθ*Sinθ)
=Csc^2_θ
得證
2016-06-30 3:44 am
left=1+tan^2 (90+θ)
left=1+1/(tan^2 θ)
left=(tan^2 θ +1)/ tan^2 θ
left=[(sin^2 θ +cos^2 θ)/cos^2 θ]/[(sin^2 θ)/cos^2 θ)]
left=1/sin^2 θ
right=tan(270^0 +θ)/[cos(180+θ)sin(360+θ)]
right=[(-1)(-1)(1)]/[(tanθ)(cosθ)(sinθ)
right=(cosθ)/[(sinθ)(cosθ)(sinθ)]
right=1/(sin^2 θ)
left=1+1/(tan^2 θ)
left=(tan^2 θ +1)/ tan^2 θ
left=[(sin^2 θ +cos^2 θ)/cos^2 θ]/[(sin^2 θ)/cos^2 θ)]
left=1/sin^2 θ
right=tan(270^0 +θ)/[cos(180+θ)sin(360+θ)]
right=[(-1)(-1)(1)]/[(tanθ)(cosθ)(sinθ)
right=(cosθ)/[(sinθ)(cosθ)(sinθ)]
right=1/(sin^2 θ)
收錄日期: 2021-04-18 15:06:44
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160611223646AA5POgu