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

问题描述:

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

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.