.[1/cos(-α)+cos(180°+ α )]/[1/sin(540°-α)+sin(360°-α)]=tan^3

来源：学生作业帮编辑：灵鹊做题网作业帮分类：数学作业时间：2024/04/29 13:23:18

.[1/cos(-α)+cos(180°+ α )]/[1/sin(540°-α)+sin(360°-α)]=tan^3α

cos(-α)=cosα,
cos(180°+ α)= -cosα
sin(540°-α)=sinα
sin(360°-α)= -sinα
所以
原式左边
=(1/cosα -cosα) / (1/sinα -sinα)
=(1-cos²α)/cosα / (1-sin²α)/sinα
=(sin²α/cosα) / (cos²α)/cosα
=(sinα/cosα)^3
=tan^3 α
这样就得到了答案

.[1/cos(-α)+cos(180°+ α )]/[1/sin(540°-α)+sin(360°-α)]=tan^3 设cos(180°+α)=1/3 且 sinα>α 求sin(180°+α)-tanα/cos(180°-α)+tan( 求证：(1/sinα-sin(180°+α))/(1/cos(540°-α)+cos(360°-α))=1/(tanα) 证明恒等式 (cosα+tanα)/[(cosα/sinα)+1/cosα]=sinα 1-sinαcos/cos平方α-sin平方=1-tanα/1+tanα 化简：tanα*(cosα-sinα)+[sinα(sinα+tanα)/1+cosα] 化简：tanα(cosα-sinα)+sinα(sinα+tanα)/1+cosα. 证明恒等式tanαsinα/tanα-sinα=1+cosα/sinα sin(540°+α)cos(360°-α)/sin(450°+α)tan(900°-α) 化简cos^2(-α)-[tan(360°+α)/sin(-α)] 化简,cos²(-α)-tan(360°+α)/sin(-α) 已知sinα+cosα=1/3,则sinαcosα+tanα-1/3cosα