tanA=(1-tanBtanC)/(tanB+tanC)
所以tanA(tanB+tanC)=(1-tanBtanC)
tanAtanB+tanCtanA=1-tanBtanC
所以tanAtanB+tanBtanC+tanCtanA=1
若A+B+C=90度,且tanA=(1-tanBtanC)/(tanB+tanC),证tanAtanB+tanBtanC+tanCtanA=1
若A+B+C=90度,且tanA=(1-tanBtanC)/(tanB+tanC),证tanAtanB+tanBtanC+tanCtanA=1
若A+B+C=90度,且tanA=(1-tanBtanC)/(tanB+tanC),证:tanAtanB+tanBtanC+tanCtanA=1
若A+B+C=90度,且tanA=(1-tanBtanC)/(tanB+tanC),证:tanAtanB+tanBtanC+tanCtanA=1
数学人气：959 ℃时间：2020-05-26 06:30:48
优质解答
我来回答
类似推荐
猜你喜欢