f(α)=[sin(π-α)cos(α-π)tan(-α+π)]/[tan(π+α)cos(3/2π+α)] 化简

化简：[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan( f(α)=[sin(π-α)cos(α-π)tan(-α+π)]/[tan(π+α)cos(3/2π+α)] 化简 ②若cos（已知α是第三象限角f(α)=sin(π-α)cos(2π-α)tan(-α-π)/tan（-α）sin（-π 化简{sin(π+α)^3cos(-α)cos(π+α)}/{tan(π+α)^3cos(-π-α)^2} 求证:[sin^2(π+α)*cos(π/2-α)+tan(2π-α)*cos(-α)]/-sin^2(-α)+tan( 若f(α)=[sin(π-α）cos(2π-α）tan(-α+π）]/[-tan(-α-π）sin（-π-α）]化简 已知f(α)=sin（π-α）cos(2π-α)tan(-α+3π/2)tan(-α-π) ／sin(-π-α).(1) α是第三象限角,f(α)={sin（α-π/2）cos（3π/2+α）tan（π-α）}/{tan（-α-π）sin（- 已知α是第三象限角,且f(α)=[sin(π+α)cos(2π-α)tan(-α+3π/2)tan(-α-π)]/sin 若α∈(0,π/6) 比较tan(sinα),tan(cosα),tan(tanα)的大小 已知tan(α+π/4)=2,则(2sinα+cosα)/(3cosα-2sinα) 已知tan(α+π/4)=2,求2cosα-sinα/cosα+3sinα