求值tan(π/6-θ)+tan(π/6+θ)+√3tan(π/6-θ)tan(π/6+θ)
问题描述:
求值tan(π/6-θ)+tan(π/6+θ)+√3tan(π/6-θ)tan(π/6+θ)
答
根号3 用两角和的正切公式展开tan(π/6-θ+π/6+θ)=a式/b式,将b式乘到等式左边再移项即得所求式子=根号3
求值tan(π/6-θ)+tan(π/6+θ)+√3tan(π/6-θ)tan(π/6+θ)
根号3 用两角和的正切公式展开tan(π/6-θ+π/6+θ)=a式/b式,将b式乘到等式左边再移项即得所求式子=根号3