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

Question

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

Answer 1

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