已知cos(π/2 +α)=3/5,求[sin(α+ 3π/2)*sin(3π/2 -α)*tan²(2π-α)*tan(π-α)][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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已知cos(π/2 +α)=3/5,求[sin(α+ 3π/2)*sin(3π/2 -α)*tan²(2π-α)*tan(π-α)][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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已知cos(π/2 +α)=3/5,求[sin(α+ 3π/2)*sin(3π/2 -α)*tan²(2π-α)*tan(π-α)][sin(α+ 3π/2)*sin(3π/2 -α)*tan²(2π-α)*tan(π-α)]÷[cos(π/2 -α)*cos(π/2 +α)]的值
已知cos(π/2 +α)=3/5,求[sin(α+ 3π/2)*sin(3π/2 -α)*tan²(2π-α)*tan(π-α)]
[sin(α+ 3π/2)*sin(3π/2 -α)*tan²(2π-α)*tan(π-α)]÷[cos(π/2 -α)*cos(π/2 +α)]的值

已知cos(π/2 +α)=3/5,求[sin(α+ 3π/2)*sin(3π/2 -α)*tan²(2π-α)*tan(π-α)][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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