tan(α+π/4)+tan(α-π/4)
=tan(α+π/4+α-π/4)[1-tan(α+π/4)tan(α-π/4)]
=tan2α[1-tan(α+π/4)tan(α+π/4-π/2)]
=tan2α[1+tan(α+π/4)cot(α+π/4)]
=tan2α[1+1]
=2tan2α
得证
化简tan（α+π/4）+tan（α-π/4）=2tan2α
1个回答
相关问题
-
化简：[ tan(π/4+α) - tan(π/4-α) ] / tan2α - tan(π/4+α)tan(π/4-α
-
化简cos(α -π )tan(α-2π)tan(2π-α)/sin(π+α)
-
tan（α+π/4）+tan（α-π/4）=2tan2α
-
化简：sin²（α+π)cos(-α+π)/tan(α+π)tan(α+2π)cos²(-α-π)
-
化简：sin2(α+π)•cos(π+α)tan(π+α)•cos3(−α−π)•tan(−α−2π)．
-
化简(2cos²α-1)/[2tan(π/4-α)sin²(π/4+α)]
-
化简2cos²α-1/2tan(π/4-α)cos²(π/4-α)
-
[sin(2π+α）·tan(-α)·cos(-α)]÷[sin(-α)·cos(2π+α）·tan(2π+α）化简
-
化简:(2cos^2α-1)/[2tan(π/4-α)sin^2(π/4+α)]
-
化简：（1）tan(π−α)•sin2(α+π2)•cos(2π−α)cos3(−π−α)•tan(α−2π)；（2）s