若(1+tanα)/(1-tanα)=2008 则1/cos2α+tan2α=

若(1+tanα)/(1-tanα)=2008 则1/cos2α+tan2α=
若(1+tanα)/(1-tanα)=2008 则1/cos2α+tan2α=

若(1+tanα)/(1-tanα)=2008 则1/cos2α+tan2α=
所求＝(sin2α+1)/cos2α=[(sinα)^2+(cosα)^2+2sinαcosα]/[(cosα)^2-(sinα)^2] (使用二倍角公式)
＝(sinα+cosα)^2/[(sinα+cosα)(cosα-sinα)
=(sinα+cosα)/(cosα-sinα)
=(1+tanα)/(1-tanα)=2008 (分子分母同除cosα)