tan(A+B)=2/54,tan(B-π)=1/4,那么tan(A+π/4)
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tan(A+B)=2/54,tan(B-π)=1/4,那么tan(A+π/4)
/>估计你的输入有误吧,4搁错地方了.
是tan(A+B)=2/5,tan(B-π/4)=1/4
tan(A+π/4)
=tan[(A+B)-(B-π/4)]
=[tan(A+B)-tan(B-π/4)]/[1+tan(A+B)tan(B-π/4)]
=(2/5-1/4)/[1+(2/5)*(1/4)]
分子分母同时乘以20
=(8-5)/(20+2)
=3/22
再问: 没有啊
再答: 啊?那为啥是2/54?
再问: 题目是这样喵怎么知道
再答: tan(A+B)=2/54=1/27, tan(B-π)=1/4, 即tanB=1/4 ∴ tanA =tan[(A+B)-B] =[tan(A+B)-tanB]/[1+tan(A+B)tanB] =[(1/27)-(1/4)]/[1+(1/27)*(1/4)] =-23/109 tan(A+π/4) =[tanA+tan(π/4)]/[1-tanAtan(π/4)] =(tanA+1)/(1-tanA) =(-23/109+1)/(1+23/109) =86/132 =43/66
是tan(A+B)=2/5,tan(B-π/4)=1/4
tan(A+π/4)
=tan[(A+B)-(B-π/4)]
=[tan(A+B)-tan(B-π/4)]/[1+tan(A+B)tan(B-π/4)]
=(2/5-1/4)/[1+(2/5)*(1/4)]
分子分母同时乘以20
=(8-5)/(20+2)
=3/22
再问: 没有啊
再答: 啊?那为啥是2/54?
再问: 题目是这样喵怎么知道
再答: tan(A+B)=2/54=1/27, tan(B-π)=1/4, 即tanB=1/4 ∴ tanA =tan[(A+B)-B] =[tan(A+B)-tanB]/[1+tan(A+B)tanB] =[(1/27)-(1/4)]/[1+(1/27)*(1/4)] =-23/109 tan(A+π/4) =[tanA+tan(π/4)]/[1-tanAtan(π/4)] =(tanA+1)/(1-tanA) =(-23/109+1)/(1+23/109) =86/132 =43/66
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