有a.maths 唔識做
回答 (3)
2008-05-22 12:25 am
2008-05-21 4:08 pm
Lets simplify tan(45-A) first.
tan(45-A)=(tan45-tanA)/(1 + tan45tanA) = (1-tanA)/(1+tanA) since tan45=1. Therefore,
L.H.S.=tanA + (1+tanA)*tan(45-A)=tanA +(1+tanA)*(1-tanA)/(1+tanA)=tanA + (1-tanA)=tanA + 1 - tanA = 1.=R.H.S.
tan(45-A)=(tan45-tanA)/(1 + tan45tanA) = (1-tanA)/(1+tanA) since tan45=1. Therefore,
L.H.S.=tanA + (1+tanA)*tan(45-A)=tanA +(1+tanA)*(1-tanA)/(1+tanA)=tanA + (1-tanA)=tanA + 1 - tanA = 1.=R.H.S.
2008-05-21 12:26 pm
參考: me
收錄日期: 2021-04-25 22:39:53
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