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

化简：[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan( 已知tanα=2,sinα+cosα＜0,求[sin(2π-α)*sin(π+α)*cos(-π+α)]/[sin(3π 求证:[sin^2(π+α)*cos(π/2-α)+tan(2π-α)*cos(-α)]/-sin^2(-α)+tan( 化简sin³（﹣α）cos(2π+α）tan（2π-α)/sin(α-2π）cos(α-3π/2)-tan(π 已知tan（3π+α）=2,求：1、（sinα+cosα）²；2、sinα-cosα/2sinα+cosα 已知tan（π-α）=2,求sinα-2sinαcosα-cosα／4cosα-3sinα的值 已知tan(α+π/4)=2,则(2sinα+cosα)/(3cosα-2sinα) 已知tan(α+π/4)=2,求2cosα-sinα/cosα+3sinα 已知tan（α+π/4）=3,求cosα+2sinα/cosα-sinα的值 化简：[sin(﹣α)cos(π－α)]∕[tan(π＋α)sin﹙3π／2＋α)] 化简{sin(π+α)^3cos(-α)cos(π+α)}/{tan(π+α)^3cos(-π-α)^2} tanα=2,求sin(π-α)cos(2π-α)sin(-α+ 3π/2)/tan(-α-π)sin(-π-α)的值