作业帮(1-tan^4θ)cos^2θ+tan^2θ
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/30 14:47:59
(1-tan^4θ)cos^2θ+tan^2θ
(1-tan^4θ)cos^2θ+tan^2θ
(1-tan^4θ)cos^2θ+tan^2θ
因为
(1-tan^4x)cos^2x+tan^2x
=(1+tan^2x)(1-tan^2x)cos^2x+tan^2x
=(1-tan^2x)[(cos^2x+sin^2x)/cos^2x]cos^2x+tan^2x
=1-tan^2x+tan^2x
=1
(1-tan^4θ)cos^2θ+tan^2θ
=(cos^2θ-sin^4θ/cos^2θ)+sin^2θ/cos^2θ
=(cos^4θ-sin^4θ)/cos^2θ+sin^2θ/cos^2θ
=(cos^4θ-sin^4θ+sin^2θ)/cos^2θ
=[cos^4θ+sin^2θ(1-sin^2θ)]/cos^2θ
=[cos^2θ(cos^2θ+sin^2θ)]/cos^2θ
=1
[1－sin^4a/cos^4a]cos²a＋tan²a={[cos^4a－sin^4a]/cos^4a}cos²a＋tan²a={[cos²a－sin²a][cos²a＋sin²a]}/cos²a＋tan²a=[cos²a－sin²a]/cos²a＋tan²a=1－sin²a/cos²a＋tan²a=1
(1-tan^4θ)cos^2θ+tan^2θ
三角比 1/tanθ-cos^2θ/tanθ
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