(1+tanα)/(1-tanα)=2008,则(1/cosα)+tan2α=?

因为(1+tanα)/(1-tanα)=(cosα +sinα)/(cosα-sinα)=2008
所以(1/cos2α)+tan2α
=(1+sin2α)/cos2α
=(sinα+cosα)^2/[(cosα)^2-(sinα)^2
=(sinα+cosα)/(cosα-sinα)
=2008