sin²θ+cos²θ=1[(m-3)/(m+5)]²+[(4-2m)/(m+5)]²=1(m-3)²+(4-2m)²=(m+5)²m²-6m+9+16-16m+4m²=m²+10m+254m²-32m-0m=0或m=8(1) m=0 ，sinθ=-3/5 与θ∈(π/2, π), 矛盾(2)m=8siinθ=5/13,cosθ=-12/13所以 m=8,tan θ=sinθ/cosθ= -5/12
