三角形ABC面积s=c^2-(a-b)^2,求c/2的正切值
问题描述：
三角形ABC面积s=c^2-(a-b)^2,求c/2的正切值
答
∵c^2=a^2+b^2-2abcosC∴S=[c^2-(a-b)^2]=(2ab-2abcosC)=2ab(1-cosC)=2ab(1-cosC)又S=absinC/22ab(1-cosC)=absinC/24ab(1-cosC)=absinC4(1-cosC)=sinC∴sinC=4(1-cosC)∴sinC/(1-cosC)=4∴(1-cosC)/sinC=1/4即tan(C/...