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2007-05-13 7:45 am
prove (1+tanA)(1+cotA)=secAcscA+2

寫埋step唔該晒...

回答 (3)

2007-05-13 9:08 pm
✔ 最佳答案
LHS = ( 1 + tanA )( 1 + cotA )
= 1 + cotA + tanA + cotAtanA
= 1 + cosA / sinA + sinA / cosA + cosAsinA / sinAcosA
= 1 + cosA / sinA + sinA / cosA + 1
= 2 + (cos^2 A + sin^2 A) / sinAcosA <--把cosA / sinA + sinA / cosA 通分母
= 2 + (1 / sinAcosA)
= 2 + ( 1/sinA ) ( 1/cosA )
= 2 + cscAsecA
= secAcscA + 2
= RHS
參考: me
(1+tanA)(1+cotA)
=1 + tanA + cotA + tanAcotA
=2 + sinA/cosA + cosA/sinA [As tanAcotA = tanA/tanA = 1]
=2 + ((sinA)^2 + (cosA)^2)/sinAcosA
=secAcscA + 2
參考: me
2007-05-13 7:50 am
(1+tanA)(1+cotA) = 1 + tanA + cotA + 1 = 2 + [(sinA)^2 + (sinA)^2] / (sinA cosA ) = 2 + 1 /(sinA cosA ) = 2+ secAcscA
參考: my knowledge


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