Question

(1/2)已知方程x^2+4ax+3a+1(a>1)的两根为tanA,tanB,且A,B属于(-派/2,派/2),则tan[(A+B)/2]的值

Answer 1

tana+tanb=-4a,tana*tanb=3a+1
tan(a+b)=-4a/(-3a)=4/3
∵a>1 ∴-4a＜0,3a＋1＞0
∴tana<0,tanb<0
a+b∈（-∏,0）
令t=tan[(a+b)/2],则
2t/(1-t²）＝4/3
t=1/2或t=-2
∴tan[(a+b)/2]=-2

Answer 2

tanA+tanB=-4a
tanAtanB=3a+1
tan(A+B)=-4a/(1-3a-1)=4/3
tan(A+B)=2tan[(A+B)/2]/{1-tan²(A+B)/2}
tan[(A+B)/2]=t
2/3=t/(1-t²)
(t+2)(2t-1)=0
t=-2 t=1/2
tan[(A+B)/2]=-2 or 1/2

Answer 3

http://zhidao.baidu.com/question/83840949.html

Answer 4

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