cotα=-1/4,则tan(π/4-α)=
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cotα=-1/4,则tan(π/4-α)=cotα=-1/4,则tan(π/4-α)=cotα=-1/4,则tan(π/4-α)=∵cotα=-1/4∴tanα=1/cotα=-4tan(π/4-α
cotα=-1/4,则tan(π/4-α)=
cotα=-1/4,则tan(π/4-α)=
cotα=-1/4,则tan(π/4-α)=
∵cotα=-1/4
∴tanα=1/cotα=-4
tan(π/4-α)=(tanπ/4-tanα)/(1+tanπ/4×tanα)
=[1-(-4)]/[1+1×(-4)]
=5/(-3)
=-5/3
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