化简[sin(π-α)cos(3π-α)tan(-α-π)]/[tan(4π-α)sin(5π+α)]
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若α∈(0,π/6) 比较tan(sinα),tan(cosα),tan(tanα)的大小
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(α+π/4)=3,求cosα+2sinα/cosα-sinα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
证明[2-2sin(α+3π/4)cos(α+π/4)]/(cos^4α-sin^4α)=(1+tanα)/(1-tan
已知tanα/2=2,求(1)tan(α-4/π)的值,(2)(3sinα+cosα)/(3cosα-sinα)
已知tanα/2=2,1.求tan(α+π/4)的值 2.(6sinα+cosα)/(3sinα-2cosα)的值
已知tanα/2=2求tan(α+π/4)的值(6sinα+cosα)/(3sinα-2cosα)的值
计算sin(-α-5π)*cos(α-2分之π)-tan(α-2分之3π)*tan(2π-α)