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[sin(45°+θ)]/[cos(45°-θ)]-[cos(θ+60°)]/[sin(θ+30°)]=?
人气:401 ℃ 时间:2020-06-03 16:46:40
解答
[sin(45°+θ)]/[cos(45°-θ)]-[cos(θ+60°)]/[sin(θ+30°)]
=[sin(45°+θ)]/[sin(45°+θ)]-[sin(θ+30°)]/[sin(θ+30°)]
=1-1
=0
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