Question

在△ABC中,a,b,c分别是角A,B,C的对边,若a²+b²=2014c²

在△ABC中,a,b,c分别是角A,B,C的对边,若a²+b²=2014c²，则（2tanA*tanB）/[tanC(tanA+tanB)]的值为

Answer 1

由余弦定理，cosC=(a²+b²-c²)/2ab=(2014c²-c²)/2ab=2013c²/2ab
由正弦定理，2013c²/2ab=2013/2*sin²C/sinAsinB
因此可以得到，cosCsinAsinB/sin²C=2013/2
于是，（2tanA*tanB）/[tanC(tanA+tanB)]
=2(sinAsinB/cosAcosB)/[(sinA/cosA+sinB/cosB)tanC]
=2(sinAsinB)/(sinAcosB+sinBcosA)tanC
=2sinAsinB/[sin(A+B)tanC]
=2sinAsinB/[sinCsinC/cosC]
=2cosCsinAsinB/sin²C
=2*2013/2=2013