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求值:tan5º+cot5º+2sec80º
如上.
人气:100 ℃ 时间:2020-09-25 05:27:19
解答
tan5º+cot5º-2sec80º
=sin5/cos5+cos5/sin5-2/sin10
=[(sin5)^2+(cos5)^2-1]/sin5cos5
=(1-1)/sin5cos5
=0
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