1+tanα／1-tanα=2006则1／cosα+tan2α=?

来源：学生作业帮编辑：灵鹊做题网作业帮分类：数学作业时间：2024/04/29 06:10:06

(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

已知(1+tan2α)/(1-tanα)=2010,求(1/cos2α)+tan2α 求证tan2分之α-1/tan2分之α=-tanα分之2 若(1+tanα)/(1-tanα)=2009,则 1/(cos2α)+tan2α=___. 证明：tanαtan2α+tan2αtan3α+……+tan(n-1)αtan(nα)＝tan(nα)/tanα-n 如果1+tanα1−tanα＝2013，那么1cos2α+tan2α=（　　）已知（1+tanα）/（1-tanα）=2010,求(1/cos2α)+tan2α的值已知tanα=1/2,tan(α-2β)=1/3,求:tan2β的值证明:sin(α+β)/cos(α-β)=tanα+tanβ/1+tanαtanβ 已知tanα=1\2 求sin2α 、cos2α、tan2α的值. 已知α∈(0,π),tanα=2,求sin4α(tan2α+1) 如果tan（45+α）=2010,那么1/cos2α+tan2α= tan(π/4+α)=2010,那么1/cos2α+tan2α=