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tan45°+cot30°/cot45°-tan60°
人气:245 ℃ 时间:2020-05-20 07:34:58
解答
(tan45°+cot30°)/(cot45°-tan60°)?
=[(tan45°+cot30°)*(sin45°*cos45°*sin30°*cos30°)]/[(cot45°-tan60°)*(sin45°*cos45°*sin30°*cos30°)]
=[(sin45°*cos30°+cos45°*sin30°)*(sin45°*cos30°)]/[(cos45°*cos60°-sin45°*sin60°)*(cos45°*sin60°)]
=[sin75°*(cos45°*sin60°)]/[cos105°*(cos45°*sin60°)]
=-tan75°
=-sin150°/(1+cos150°)
=-sin30°/(1-cos30°)
=-(1/2)/(1-√3 / 2)
=-√3 - 2
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