sinA=(m-3)/(m+5)cosA=(4-2m)/(m+5)A∈(π/2,π)则tanA=
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sinA=(m-3)/(m+5)cosA=(4-2m)/(m+5)A∈(π/2,π)则tanA=sinA=(m-3)/(m+5)cosA=(4-2m)/(m+5)A∈(π/2,π)则tanA=sinA
sinA=(m-3)/(m+5)cosA=(4-2m)/(m+5)A∈(π/2,π)则tanA=
sinA=(m-3)/(m+5)cosA=(4-2m)/(m+5)
A∈(π/2,π)
则tanA=
sinA=(m-3)/(m+5)cosA=(4-2m)/(m+5)A∈(π/2,π)则tanA=
∵sinA=(m-3)/(m+5),cosA=(4-2m)/(m+5)
又sin²A+cos²A=1
∴(m-3)²/(m+5)²+(4-2m)²/(m+5)²=1
∴5m²-22m+25=m²+10m+25
∴4m²-32m=0
解得m=0或m=8
m=0时,sinA=-3/5与A∈(π/2,π)矛盾
m=8时,sinA=5/13,cosA=-12/13
∴tanA=sinA/cosA=-5/12
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