1+tanα／1-tanα=2006则1／cosα+tan2α=?

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(1+tanα)／(1-tanα)=2006
1+tanα=2006(1-tanα)
tanα=2005/2007
cos2α=(1-tan^2α)/(1+tan^2α)
=(1-2005^2/2007^2)/(1+2005^2/2007^2)
tan2α=2tanα/(1-tan^2α)
=(2*2005/2007)/(1-2005^2/2007^2)
1／cos2α+tan2α
=1/[(1-2005^2/2007^2)/(1+2005^2/2007^2)]+(2*2005/2007)/(1-2005^2/2007^2)
=(1+2005^2/2007^2)/(1-2005^2/2007^2)+(2*2005/2007)/(1-2005^2/2007^2)
=(1+2005^2/2007^2+2*2005/2007)/(1-2005^2/2007^2)
=(1+2005/2007)^2/[(1-2005/2007)(1+2005/2007)]
=(1+2005/2007)/(1-2005/2007)
=(4012/2007)/(2/2007)
=4012/2
=2006
1／cos2α+tan2α
=1/[(1-tan^2α)/(1+tan^2α)]+2tanα/(1-tan^2α)
=(1+tan^2α)/(1-tan^2α)+2tanα/(1-tan^2α)
=(1+tan^2α+2tanα)/(1-tan^2α)
=(1+tanα)^2/(1-tan^2α)
=(1+tanα)^2/[(1-tanα)(1+tanα)]
=(1+tanα)/(1-tanα)
=(1+tanα)/(1-tanα)
=2006

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