已知tan(2A-B)=√3(1+M),tan(A-B)-√3[M-tan(2A-B)tan(A-B)],求tanA的值_数学解答

已知tan(2A-B)=√3(1+M),tan(A-B)-√3[M-tan(2A-B)tan(A-B)],求tanA的值

人气：418 ℃　时间：2020-03-07 12:48:15

解答

tanA
=tan[(2A-B)--(A-B)]
=[ tan(2A-B)--tan(A-B) ] / [1+tan(2A-B)tan(A-B)]
=[√3(1+M)--tan(A-B)] / tan(A-B)-√3[M-tan(2A-B)tan(A-B)],
=1