Question

化简：sin(π-α)cos[(3π/2)-α]tan(π-α)/sin(2π-α)cos(π-α)tan(3π+α)要求详细过程

Answer 1

{sin(π-α)cos[(3π/2)-α]tan(π-α)}/{sin(2π-α)cos(π-α)tan(3π+α)}
=[sinα(-sinα)(-tanα)]/[-sinα(-cosα)*tanα]
=sinα/cosα
=tanα
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