化简cos(α+π)tan(2π+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(
f(α)=[sin(π-α)cos(α-π)tan(-α+π)]/[tan(π+α)cos(3/2π+α)] 化简
化简sin(α-2π)cos(3π/2-α)+tan(π-α)tan(π/2+α)+cos^2(π-α)
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
化简:tan(π-α)·sin²(α+π/2)·cos(2π-α)/cos³(-α-π)·tan(α
化简:(1)、tan(2π+α)tan(π+α)cos(-π-α)的3次方分之sin(-α)的平方乘以cos(π+α);
化简(2cosα^2-1/cosα^2+tanα^2)/(2tan(π/4-α)sin(π/4+α)^2
化简sin^2(α+π)cos(π+α)cot(-α+2π)/tan(π-α)cos^3(-α-π)
已知cos(2α+β)=3cosβ,化简tanα×tan(α+β)