化简(2cosα^2-1/cosα^2+tanα^2)/(2tan(π/4-α)sin(π/4+α)^2
化简:[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan(
化简(sinα)^2tanα+cosα+(cosα)^2*1/tanα+2sinαcosα
化简:(sin²αtanα+cos²α/tanα+2sinαcosα)sinαcosα
求证:(1-sinα+cosα)/(1+sinα+cosα)=tan(π/4-α/2)
求证:1-2sinαcosα/cos²α-sin²α=tan(π/4-α)
试证明:1+2sinαcosα/cos平方α-sin平方α=tan(π/4-α)
求证(1-2sinαcosα)/(cos²α-sin²α)=tan(π/4-α)
已知tanα/2=2,求(1)tan(α-4/π)的值,(2)(3sinα+cosα)/(3cosα-sinα)
化简(2cosα^2-1/cosα^2+tanα^2)/(2tan(π/4-α)sin(π/4+α)^2
证明[2-2sin(α+3π/4)cos(α+π/4)]/(cos^4α-sin^4α)=(1+tanα)/(1-tan
求证:[sin^2(π+α)*cos(π/2-α)+tan(2π-α)*cos(-α)]/-sin^2(-α)+tan(
化简:[cos(α/2)+sin(α/2)][cos(α/2)-sin(α/2)][1+tanαtan(α/2)]