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