·若tan(π/4-α)=3,则cotα等于
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·若tan(π/4-α)=3,则cotα等于·若tan(π/4-α)=3,则cotα等于·若tan(π/4-α)=3,则cotα等于tan-α=tan[(π|4-α)-π|4]=[tan(π|4-α)
·若tan(π/4-α)=3,则cotα等于
·若tan(π/4-α)=3,则cotα等于
·若tan(π/4-α)=3,则cotα等于
tan-α=tan[(π|4-α)-π|4]=[tan(π|4-α)-tan(π|4)]/1+tan(π|4-α)
=1/2
所以tanα=-1/2
cotα等于1/tanα=-2
tan(π/4-α)
=(tanπ/4-tanα)/(1+tanπ/4tanα)
=(1-tanα)/(1+tanα)=3
1-tanα=3+3tanα
tanα=-1/2
所以cotα=1/tanα=-2
tan(π/4-α)
=(tanπ/4-tanα)/(1+tanπ/4tanα)
=(1-tanα)/(1+tanα)=3
1+tanα不等于0
则
1-tanα=3+3tanα
tanα=-1/2
所以
cotα=1/tanα=-2
·若tan(π/4-α)=3,则cotα等于
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