f(α)=sin(π-α)*cos(2π-α)/sin(3π/2-α)tanα,则f(-10π/3)等于

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 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 f(α)=[sin(π-α)cos(α-π)tan(-α+π)]/[tan(π+α)cos(3/2π+α)] 化简 已知f(α)=[sin(π-α)cos(2π-α)tan(-α+3π/2)]/[cot(-π-α)sin(-π-α)] 化简 f（α）=sin(π—α) cos(2π—α)tan（-α+3/2π）/cot(-α-π）sin(-π-α) 已知α为第三象限角,且f(α)=sin(π-α)cos(2π-α)tan(-α+π)/sin(π+α)tan(2π-α) 已知f(α)=sin(π-α)cos(2π-α)tan(π-α)/-tan(-π-α)sin(-π-α),若α是第三象限 若f(α)=[sin(π-α）cos(2π-α）tan(-α+π）]/[-tan(-α-π）sin（-π-α）]化简 有关三角函数的题已知f(α)=sin(π-α)cos(2π-α)tan(-α+π)/-tan(-α-π)sin(-π-α 已知tan(α+π/4)=2,则(2sinα+cosα)/(3cosα-2sinα)