在三角形ABC中,a,b,c分别是角A,B,C的对边,若a^2+b^2=2014c^2,则（2tanA×tanB)/ta
在三角形ABC中,a,b,c分别是角A,B,C的对边,若a^2+b^2=2014c^2,则（2tanA×tanB)/tanC(tanA+tanB)的值为
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cosC=(a²+b²-c²)/2ab
=(2014c²-c²)/2ab
=2013c²/2ab
由正弦定理
=2013/2*sin²C/sinAsinB
∴cosCsinAsinB/sin²C=2013/2
原式=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
=(2014c²-c²)/2ab
=2013c²/2ab
由正弦定理
=2013/2*sin²C/sinAsinB
∴cosCsinAsinB/sin²C=2013/2
原式=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
1年前
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