【sin(α+2kπ)+cos(π/2+α)+tan(3π-α)】/【sin(α-π)+cos(α-π/2)+cot(π/2-α)】
问题描述:
【sin(α+2kπ)+cos(π/2+α)+tan(3π-α)】/【sin(α-π)+cos(α-π/2)+cot(π/2-α)】
答
【sin(α+2kπ)+cos(π/2+α)+tan(3π-α)】/【sin(α-π)+cos(α-π/2)+cot(π/2-α)】
=(sina-sina-tana)/(-sina+sina+tana)=-1