解：原式={cos[-(π-α)]cos[4π+(π-α)]}/{sin[-(π-2α)]tan[2π+(π-α)]}
=[cos(π-α)cos(π-α)]/[-sin(π-2α)tan(π-α)]
=[cos²(π-α)]/[-sin2α×(-tanα)]
=(-cosα)²/[-sin2α×(-sinα/cosα)]
=cos²α/[-sinαcosα×(-sinα/cosα)]
=cos²α/sin²α
=cot²α
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