化简sin(α-2π)cos(3π/2-α)+tan(π-α)tan(π/2+α)+cos^2(π-α)

化简：[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan( 求证:[sin^2(π+α)*cos(π/2-α)+tan(2π-α)*cos(-α)]/-sin^2(-α)+tan( 化简{sin(π+α)^3cos(-α)cos(π+α)}/{tan(π+α)^3cos(-π-α)^2} 化简sin³（﹣α）cos(2π+α）tan（2π-α)/sin(α-2π）cos(α-3π/2)-tan(π 化简sin^2(α+π)cos(π+α)cot(-α+2π)/tan(π-α)cos^3(-α-π) 已知tan(α+π/4)=2,则(2sinα+cosα)/(3cosα-2sinα) 已知tan(α+π/4)=2,求2cosα-sinα/cosα+3sinα f(α)=[sin(π-α)cos(α-π)tan(-α+π)]/[tan(π+α)cos(3/2π+α)] 化简 化简sin(α-2π)cos(3π/2-α)+tan(π-α)tan(π/2+α)+cos^2(π-α) 已知tan（α+π/4）=3,求cosα+2sinα/cosα-sinα的值 tan(π-α)*sin^2(α+π/2)*cos(2π-α)/cos^3(-α-π)*tan(2π-α) sin^2就 [sin(2π+α）·tan(-α)·cos(-α)]÷[sin(-α)·cos(2π+α）·tan(2π+α）化简