Find tan(A-B) if tanA and tanB are roots of x^2 - (1+k)x +k = 0, 0<B<A<90?
2020-12-22 3:36 pm
I have found with tanA+tanB = 1+k, tanAtanB=k, tan(A+B)=(1+k)/(1-k), but unable to find tan(A-B). Please help.
回答 (1)
2020-12-22 4:25 pm
✔ 最佳答案
0 < B < A < 90°故 0 < tanB < tanA < +∞.
k = tanA．tanB > 0
解方程式:
0 = x^2 - (1+k)x + k = (x-1)(x-k)
或直接套公式:
tanA = {(1+k)+√[(1+k)^2-4k]}/2
= {(1+k)+|1-k|}/2 = max{1,k}
tanB = {(1+k)-|1-k|}/2 = min{1,k}
tan(A-B) = (tanA -tanB)/(1+tanA tanB)
= (max{1,k}-min{1,k})/(1+k)
= |1-k|/(1+k)
收錄日期: 2021-04-24 08:16:40
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20201222073618AAV8z4z