若tanα=-1/3,tan(β-(π/4))=-1/3,则tan(α+β)=
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若tanα=-1/3,tan(β-(π/4))=-1/3,则tan(α+β)=
若tanα=-1/3,tan(β-(π/4))=-1/3,则tan(α+β)=
若tanα=-1/3,tan(β-(π/4))=-1/3,则tan(α+β)=
tanα=-1/3 tan(β-π/4)=-1/3
tan(α+β-π/4)=[tanα+tan(β-π/4)]/[1-tanα·tan(β-π/4)]
=[(-1/3)+(-1/3)]/[1-(-1/3)(-1/3)]
=-3/4
tan(α+β)
=tan[(α+β-π/4)+π/4]
=[tan(α+β-π/4)+tan(π/4)]/[1-tan(α+β-π/4)·tan(π/4)]
=[(-3/4)+1]/[1-(-3/4)×1]
=1/7
若tan(α+β)=3/5,tan(β-(π)/4)=1/4,则tan(α+(π)/4)=?
若tanα=-1/3,tan(β-(π/4))=-1/3,则tan(α+β)=
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