已知sinα是方程5X²-7X-6=0的根，
求[sin(α+3π/2)]×sin(3π/2-α)×tan²（2π-α)×tan(π-α)]÷[cos(π/2-α)×cos(π/2+α)]的值...
求[sin(α+3π/2)]×sin(3π/2-α)×tan²（2π-α)×tan(π-α)]÷[cos(π/2-α)×cos(π/2+α)]的值
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5X²-7X-6=0 , (5x+3)(x-2)=0 , sina=-3/5 , cosa=±4/5
[sin(α+3π/2)]×sin(3π/2-α)×tan²（2π-α)×tan(π-α)]÷[cos(π/2-α)×cos(π/2+α)]
=-cos²axtan²ax(-tana)÷(-sin²a)
=-tana
=±3/4
[sin(α+3π/2)]×sin(3π/2-α)×tan²（2π-α)×tan(π-α)]÷[cos(π/2-α)×cos(π/2+α)]
=-cos²axtan²ax(-tana)÷(-sin²a)
=-tana
=±3/4
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很简单那5X²-7X-6=0 , (5x+3)(x-2)=0 , sina=-3/5 , cosa=±4/5
[sin(α+3π/2)]×sin(3π/2-α)×tan²（2π-α)×tan(π-α)]÷[cos(π/2-α)×cos(π/2+α)]
=-cos²axtan²ax(-tana)÷(-sin²a)
=-tana
=±3/4
[sin(α+3π/2)]×sin(3π/2-α)×tan²（2π-α)×tan(π-α)]÷[cos(π/2-α)×cos(π/2+α)]
=-cos²axtan²ax(-tana)÷(-sin²a)
=-tana
=±3/4
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