已知sinx=(m-3)/(m+5),cosx=(4-2m)/(m+5)(π/2<x<π),则tanx=

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已知sinx=(m-3)/(m+5),cosx=(4-2m)/(m+5)(π/2<x<π),则tanx=

已知sinx=(m-3)/(m+5),cosx=(4-2m)/(m+5)(π/2<x<π),则tanx=
已知sinx=(m-3)/(m+5),cosx=(4-2m)/(m+5)(π/2<x<π),则tanx=

已知sinx=(m-3)/(m+5),cosx=(4-2m)/(m+5)(π/2<x<π),则tanx=
tanx
=sinx/cosx
=[(m-3)/(m+5)] / [(4-2m)/(m+5)]
=(m-3)/(4-2m)
=(m-3)/2(2-m)

利用(sinx)^2+(cosx)^2=1
所以[(m-3)^2+(4-2m)^2]/(m+5)^2=1
5m^2-22m+25=m^2+10m+25
所以m=0或m=8
x∈(π/2,π),sinx>0
所以m=8
sinx=5/13
cosx=-12/13
tanx=-5/12