由正弦定理得
a/sinA=b/sinB=c/sinC,又a²、b²、c²成等差數(shù)列,因此
2sin²B=sin²A+sin²C
1-cos(2B) =[1-cos(2A)]/2 +[1-cos(2C)]/2
cos(2A)+cos(2C)=2cos(2B)=2cos(π/2)=0
2cos(A+C)cos(A-C)=0
-2cosBcos(A-C)=0
cos(A-C)=0
A-C=π/2(C>0,0C=A+π/2
A=π-B-C=π-π/4-(A+π/2)
2A=π/4
A=π/8
tan(2A)=tan(π/4)=1=2tanA/(1-tan²A)
整理,得
tan²A+2tanA=1
(tanA+1)²=2
tanA=-√2-1(00,舍去)或tanA=√2-1

tanA=√2-1