化简[sin(π-α)tan(5π/2+α)]/[cos(2π-α)cot(π/2-α)]分后给

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/29 08:58:30
化简[sin(π-α)tan(5π/2+α)]/[cos(2π-α)cot(π/2-α)]分后给
x){3< 6j$io7rb5 R% ;iG }wϴI*'nYv6Tsα<(snc=@68N}X P>0W I */.H̳)AW

化简[sin(π-α)tan(5π/2+α)]/[cos(2π-α)cot(π/2-α)]分后给
化简[sin(π-α)tan(5π/2+α)]/[cos(2π-α)cot(π/2-α)]
分后给

化简[sin(π-α)tan(5π/2+α)]/[cos(2π-α)cot(π/2-α)]分后给
[sin(π-α)tan(5π/2+α)]/[cos(2π-α)cot(π/2-α)]
=-[sinαcotα]/(cosαtanα)
=-sinα/cosα*(cotα/tanα)
=-tanα*cotα/tanα
=-cotα

cotα

sin(π-α)tan(5π/2+α)]/[cos(2π-α)cot(π/2-α)]