若sinθ=(m-3)/(m+5),cosθ=(4-2m)/(m+5),θ∈(π/2,π),则m= ,tanθ =

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若sinθ=(m-3)/(m+5),cosθ=(4-2m)/(m+5),θ∈(π/2,π),则m=,tanθ=若sinθ=(m-3)/(m+5),cosθ=(4-2m)/(m+5),θ∈(π/2,π)

若sinθ=(m-3)/(m+5),cosθ=(4-2m)/(m+5),θ∈(π/2,π),则m= ,tanθ =
若sinθ=(m-3)/(m+5),cosθ=(4-2m)/(m+5),θ∈(π/2,π),则m= ,tanθ =

若sinθ=(m-3)/(m+5),cosθ=(4-2m)/(m+5),θ∈(π/2,π),则m= ,tanθ =
sin²θ+cos²θ=1
[(m-3)/(m+5)]²+[(4-2m)/(m+5)]²=1
(m-3)²+(4-2m)²=(m+5)²
m²-6m+9+16-16m+4m²=m²+10m+25
4m²-32m-0
m=0或m=8
(1) m=0 ,sinθ=-3/5 与θ∈(π/2, π), 矛盾
(2)m=8
siinθ=5/13,cosθ=-12/13
所以 m=8,tan θ=sinθ/cosθ= -5/12