化简:{sin^2(α+π)*cos(α+π)*cos(-α-2π)}/{sin(-α-2π)*tan(π+α)*cos
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
化简:{sin^2(α+π)*cos(α+π)*cos(-α-2π)}/{sin(-α-2π)*tan(π+α)*cos
求证:(1-sinα+cosα)/(1+sinα+cosα)=tan(π/4-α/2)
求证:1-2sinαcosα/cos²α-sin²α=tan(π/4-α)
已知tan(α+π/4)=3,求cosα+2sinα/cosα-sinα的值
试证明:1+2sinαcosα/cos平方α-sin平方α=tan(π/4-α)
化简:[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan(
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(
化简sin³(﹣α)cos(2π+α)tan(2π-α)/sin(α-2π)cos(α-3π/2)-tan(π
[sin(2π+α)·tan(-α)·cos(-α)]÷[sin(-α)·cos(2π+α)·tan(2π+α)化简