tan(a+b)=2/5 tan(a+π/4)=3/22 求tan(b-π/4)
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tan(a+b)=2/5 tan(a+π/4)=3/22 求tan(b-π/4)
tan(a+b)=2/5 tan(a+π/4)=3/22 求tan(b-π/4)
tan(a+b)=2/5 tan(a+π/4)=3/22 求tan(b-π/4)
tan(a+b)=tan[(a+π/4)+(b-π/4)]=tan(a+π/4)+tan(b-π/4)/1-tan(a+π/4)×tan(b-π/4)=2/5 再将 tan(a+π/4)=3/22 带入可解得tan(b-π/4) =1/4 本题考点在于灵活地变换结构从而简便的做出解答!
因为tana(a+b)=tan[(a+兀/4)+(b-兀/4)]=[tan(a+兀/4)+tan(b-兀/4)]/[1-tan(a+兀/4)*tan(b-兀/4)]=[3/22+tan(b-兀/4)]/[1-(3/22)*tan(b-兀/4)]=2/5,由此解得tan(b-兀/4)=1/4
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tan(a+b)=2/5 tan(a+π/4)=3/22 求tan(b-π/4)
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