若1+tanx/1-tanx=2005,则,1/cos2x+tan2x=?
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若1+tanx/1-tanx=2005,则,1/cos2x+tan2x=?
1/cos2x
=1/[(cosx)^2-(sinx)^2]
=[(cosx)^2+(sinx)^2]/[(cosx)^2-(sinx)^2]
=[1+(tanx)^2]/[1-(tanx)^2]
tan2x = 2tanx /[1 - (tanx)^2]
1/cos2x+tan2x
=[1+(tanx)^2]/[1-(tanx)^2]+2tanx /[1 - (tanx)^2]
=[1+(tanx)^2+2tanx] /[1 - (tanx)^2]
=(1+tanx)^2/(1+tanx)(1-tanx)
=(1+tanx)/(1-tanx)
=2005
=1/[(cosx)^2-(sinx)^2]
=[(cosx)^2+(sinx)^2]/[(cosx)^2-(sinx)^2]
=[1+(tanx)^2]/[1-(tanx)^2]
tan2x = 2tanx /[1 - (tanx)^2]
1/cos2x+tan2x
=[1+(tanx)^2]/[1-(tanx)^2]+2tanx /[1 - (tanx)^2]
=[1+(tanx)^2+2tanx] /[1 - (tanx)^2]
=(1+tanx)^2/(1+tanx)(1-tanx)
=(1+tanx)/(1-tanx)
=2005
一直1+tanx/1-tanx=2012则1/cos2x+tan2x=?
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