求证: sin^2/(sin-cos) - (sin+cos)/(tan^2 -1) =sin+cos 1个回答 #热议# 不吃早饭真的会得胆结石吗? asd20060324 2011-03-27 · TA获得超过5.4万个赞 知道大有可为答主 回答量:1.8万 采纳率:62% 帮助的人:8666万 我也去答题访问个人页 关注 展开全部 sin^2/(sin-cos) - (sin+cos)/(tan^2 -1) =sin^2/(sin-cos) -(sin+cos)/[(sin^2/cos^2)-1]=sin^2/(sin-cos) -(sin+cos)cos^2/(sin^2-cos^2)=sin^2/(sin-cos) -cos^2/(sin-cos)(sin^2-cos^2)/(sin-cos)=sin-cos 本回答由提问者推荐 已赞过 已踩过< 你对这个回答的评价是? 评论 收起 推荐律师服务: 若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询 其他类似问题 2022-09-10 求证:(1-sin 2α)/(cos^2 α-sin^2 α)=(1-tan α)/(1+tan α) 2012-12-07 cos(2π-α)sin(π+α)/sin(π/2+α)tan(3π-α) 2 2012-11-23 已知sin(540°+ɑ)=-1/3,求[sin(180°+ɑ)*cos(720°+α)*tan(540°+α)]/[cos(-α-180°)* 3 2010-12-09 已知sin^2α/sin^2β+cos^2αcos^2γ=1,求证:tan^2α/tan^2β=sin^2γ,非常感谢! 3 2011-02-25 sin²π/3+cos^43π/2-tan²π/3 3 2012-06-24 求值sin^2(17)+cos^2(47)+sin17cos47 急 sinx=1/3,2π<a<3π则sina/2+cosa/2= 3 2011-03-29 1.求值sin(-4/5π)cos(-5/3π)-tan(-11/6π)-(cos7/6 π×3tan4/3 π)/sin2/3 π 4 2011-03-03 已知sin(π+α)=-1/2 计算 cos(2π-α) tan(α-7π) 3 为你推荐: