已知2tanA=3tanB。求证:tan(A-B)=sin2B/5-cos2B
1个回答
展开全部
tan(A-B)=(tanA-tanB)/(1+tanAtanB)
=(tanB/2)/(1+3tan^2B/2)
=tanB/(2+3tan^2B)
=sinBcosB/(2cos^2B+3sin^2B)
=2sinBcosB/(4+2sin^2B)
=sin2B/(4+1-cos2B)
=sin2B/(5-cos2B)
=(tanB/2)/(1+3tan^2B/2)
=tanB/(2+3tan^2B)
=sinBcosB/(2cos^2B+3sin^2B)
=2sinBcosB/(4+2sin^2B)
=sin2B/(4+1-cos2B)
=sin2B/(5-cos2B)
本回答由提问者推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
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
