tan(π/4+α)=2010,那么1/cos2α+tan2α=

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tan(π/4+α)=2010,那么1/cos2α+tan2α=tan(π/4+α)=2010,那么1/cos2α+tan2α=tan(π/4+α)=2010,那么1/cos2α+tan2α=利用三角

tan(π/4+α)=2010,那么1/cos2α+tan2α=
tan(π/4+α)=2010,那么1/cos2α+tan2α=

tan(π/4+α)=2010,那么1/cos2α+tan2α=
利用三角万能公式可得
1/cos2α+tan2α
=(1+tan^2α)/(1-tan^2α)+2tanα/(1-tan^2α)
=(1+tanα)^2/[(1-tanα)(1+tanα)]
=(1+tanα)/(1-tanα)
=(tanπ/4+tanα)/(1-taπ/4tanα)
=tan(π/4+α)
=2010