化简:根号下1-cos(2π+θ)/1+cos(2π+θ)+根号下1+cos(2π+θ)/1-cos(2π+θ)( π
问题描述:
化简:根号下1-cos(2π+θ)/1+cos(2π+θ)+根号下1+cos(2π+θ)/1-cos(2π+θ)( π
答
√[1-cos(2π+θ)]/[1+cos(2π+θ)]+√[[1+cos(2π+θ)]/[1-cos(2π+θ)]=√(1-cosθ)/(1+cosθ)+√(1+cosθ)/(1-cosθ)=√(1-cosθ)^2/(sinθ)^2+√(1+cosθ)^2/(sinθ)^2=|1-cosθ|/|sinθ|+|1+cosθ|/|sinθ|=[(1...答案好像是-2/sinθ?...-2cscθ=-2/sinθ