问题描述: 已知(1+tan2α)/(1-tanα)=2010,求1/cos2α+tan2α的值 1个回答 分类:数学 2014-10-04 问题解答: 我来补答 (1+tan2α)/(1-tanα)=2010=>{1+2tanα/[(1-tanα)^2]}/(1-tanα)=1-(tanα)^2+2tanα=2010(1+tanα)=>2009+(tanα)^2+2008tanα=0 (1)=>(1+tanα)=[tanα-(tanα)^2]/20091/cos2α+tan2α=(1+sin2α)/cos2α=(1+tanα)/(1-tanα)=tanα/2009根据(1)可求出tanα~=-1,-2007,就可求出上式~=-1/2009 或者-2007/2009. 展开全文阅读