化简sin(π/2+α)cos(3π-α)tan(π+α)/cos(π/2-α)cos(-α-π)
问题描述:
化简sin(π/2+α)cos(3π-α)tan(π+α)/cos(π/2-α)cos(-α-π)
答
sin(π/2+α)cos(3π-α)tan(π+α)/cos(π/2-α)cos(-α-π)
=[cosα*(-cosα)*tanα]/[sinα*(-cosα)]
=[-sinα*cosα]/[sinα*(-cosα)]
=1.