find all the triangles for ABC for which tan(A-B)+tan(B-C)+tan(C-A)=0
2006-11-08 20:51:04 · 2 answers · asked by missy2589 2 in Science & Mathematics ➔ Mathematics
Where it sais tamA-tanB it is meant to say tanA-tanB sorry typo!
2006-11-08 20:53:30 · update #1
tan(A-B)+tan(B-C)+tan(C-A)
= (tanA - tanB)/(1 + tanAtanB) + (tanB - tanC)/(1 + tanBtanC) + (tanC - tanA)/(1 + tanAtanC)
= [(tanA - tanB)(1 + tanBtanC)(1 + tanAtanC) + (tanB - tanC)(1 + tanAtanC)(1 + tanAtanB) + (tanC - tanA)(1 + tanAtanB)(1 + tanBtanC)]/[(1 + tanAtanB)(1 + tanBtanC)(1 + tanAtanC)]
= 0
ie (tanA - tanB)(1 + tanBtanC)(1 + tanAtanC) + (tanB - tanC)(1 + tanAtanC)(1 + tanAtanB) + (tanC - tanA)(1 + tanAtanB)(1 + tanBtanC) = 0
On expanding
(tanA - tanB)(1 + tanAtanC + tanBtanC + tanAtanBtan²C)
+ (tanB - tanC)(1 + tanAtanC + tanAtanB + tan²AtanBtanC)
+ (tanC - tanA)(1 + tanAtanB + tanBtanC + tanAtan²BtanC)
= tanA - tanB + tan²AtanC - tanAtanBtanC + tanAtanBtanC - tan²BtanC + tan²AtanBtan²C - tanAtan²Btan²C
+ tanB - tanC + tanAtanBtanC - tanAtan²C + tanAtan²B - tanAtanBtanC + tan²Atan²BtanC - tan²AtanBtan²C
+ tanC - tanA + tanAtanBtanC - tan²AtanB + tanBtan²C - tanAtanBtanC + tanAtan²Btan²C - tan²Atan²BtanC
= tan²AtanC - tan²BtanC - tanAtan²C + tanAtan²B - tan²AtanB + tanBtan²C
=0
Divide all by tanAtanBtanC
So tanA/tanB - tanB/tanA - tanC/tanB + tanB/tanC - tanA/tanC - tanC/tanA = 0
ie tanA/tanB + tanB/tanC + tanC/tanA = tanB/tanA + tanC/tanB = tanA/tanC
Whence tanA/tanB + tanB/tanC + tanC/tanA = cotA/cotB + cotB/cotC + cotC/cotA
Since a triangle then tan(A + B) = tan(π - C) = - tanC
So (tanA + tanB)/(1 - tanAtanB) = -tanC
ie tan A + tan B = tanAtanBtanC - tanC
so tanA + tanB + tanC = tanAtanBtanC
2006-11-08 21:47:43 · answer #1 · answered by Wal C 6 · 0⤊ 0⤋
just expand all three u get
1+tanA,1+tanB,1+tanC in the denominatorseperately just take lcm so than expand it throughly u will get the required answer.i have tried it here just u 2 try it ok.
2006-11-09 05:39:56 · answer #2 · answered by ambresh 2 · 0⤊ 0⤋