求证：[tan（2π-α）cos（3π/2-α)cos(6π-α)]/[sin(α+3π/2)cos(α+3π/2)]

证明：左边=[-tanα*cos（-π/2-α)cos(-α)]/[sin(α-π/2)cos(α-π/2)]
=[-tanα*cos（π/2+α)*cosα]/[-sin(π/2-α)cos(π/2 -α)]
=[tanα*(-sinα)*cosα]/(cosα*sinα)
=-tanα
=右边
等式得证.