作业帮1)tan(3π-α)/sin(π-α)sin(3/2π-α)+sin(2π-α)cos(α-7/2π)/sin(3/2π+α)cos(2π+α)
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/30 05:59:12
1)tan(3π-α)/sin(π-α)sin(3/2π-α)+sin(2π-α)cos(α-7/2π)/sin(3/2π+α)cos(2π+α)
1)tan(3π-α)/sin(π-α)sin(3/2π-α)+sin(2π-α)cos(α-7/2π)/sin(3/2π+α)cos(2π+α)
1)tan(3π-α)/sin(π-α)sin(3/2π-α)+sin(2π-α)cos(α-7/2π)/sin(3/2π+α)cos(2π+α)
原式=-tanα/sinα(-cosα)-sinα(-sinα)/(-cosα)cosα
=(sinα/cosα)/sinαcosα-sin²α/cos²α
=(1-sin²α)/cos²α
=1
tan(3π-α)/sin(π-α)sin(3/2π-α)+sin(2π-α)cos(α-7/2π)/sin(3/2π+α)cos(2π+α)
=tan(-α)/(-sinαcosα)-sin(-α)sinα/(-cosα)cosα
=tanα/(sinαcosα)-sin^2α/cos^α
=1/cos^2α-sin^2α/cos^2α
=cos^2α/cos^2...
全部展开
tan(3π-α)/sin(π-α)sin(3/2π-α)+sin(2π-α)cos(α-7/2π)/sin(3/2π+α)cos(2π+α)
=tan(-α)/(-sinαcosα)-sin(-α)sinα/(-cosα)cosα
=tanα/(sinαcosα)-sin^2α/cos^α
=1/cos^2α-sin^2α/cos^2α
=cos^2α/cos^2α
=1
很高兴为您解答，祝你学习进步！【学习宝典】团队为您答题。
有不明白的可以追问！如果您认可我的回答。
请点击下面的【选为满意回答】按钮，谢谢！
收起