化简tan(2π-α)sin(-2π-α)cos(6π-α)/cos(α-π)sin(5π-α)

若a属于(0,π/2)试比较tanα、tan(tanα)、tan(sinα) 化简cos2(-α)-tan(2π+α)/sin(-α) 化简cos(α -π )tan(α-2π)tan(2π-α)/sin(π+α) cos(π-α)tan(2π-α)tan(π-α)/sin(π+α) 化简 化简sin(2π-α）tan（α+π）tan（-α-π）/ 化简:cos(2π-α)sin(-π-α)/sin(π/2+α)tan(3π-α) 1.求证tanαsinα/tanα-sinα=tanα+sinα/tanαsinα2.已知sinα+cosα=-1/5(π/2 化简[sin²(α+π)cos(α+π)]/tan(α+π)cos³(-α-π)tan(-α-2π)化简sin²(α+π)cos(α+π)/tan(α+π)cos³(-α-π)tan(-α-2π) sin(2π-α)tan(α+3π)tan(-α-π)/tan(α-3π)cos(π-α) 化简cos(α-π)sin²(α+2π)/tan (α+4π)tan(α+π)cos³(-α-π) 化简:(sin(α-3π)*cos(-α))/(tan(α-5π)*cos^2(α-5π)*tan(4π-α)) 化简：(sin^2α-tan^2α)/(sin^2α·tan^2α) 化简：[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan(α-3π)] 已知tanα-tanβ=2tanα^2tanβ,且αβ均不等于kπ/2.试求sinβsin(2α+β）/sinβ或者能给思路已知就是这个：tanα-tanβ=2tanαtanαtanβ 已知f(α)=(sin(Π／2-α)cos(2Π-α)tan(-α+3Π))／(tan(Π+α)sin(Π／2＋α)),化简f(α) 化简sin(9π+α）tan（3/2π+α）/cos(-4π-α）tan（15π+α）tan(7/2π-α） 若f(α)=[sin(π-α）cos(2π-α）tan(-α+π）]/[-tan(-α-π）sin（-π-α）]化简 化简sin(3π-α)/[sin(-α-2π)]+cos(4π-α)/[cos(α-5π)]+tan(-α)/[tan(π+α)]