S=c^2-(a-b)^2,
而,S=1/2ab*sinC=c^2-a^2-b^2+2ab,
cosC=(a^2+b^2-c^2)/2ab,
c^2-a^2-b^2=-2ab*cosC,
即有,
1/2*ab*sinC=-2ab*cosC+2ab,
sinC=4-4cosC,
又因?yàn)?tanC=2tan(C/2)/[1-tan^2(C/2)],
sinC=2tan(C/2)/[1+tan^2(C/2),
cosC=[1-tan^2(C/2)]/[1+tan^2(C/2)],
令,tan(C/2)=m,則有,
sinC=4-4cosC,
2m/(1+m^2)=4-4*(1-m^2)/(1+m^2),
解得m=1/4,
m=1/4=tan(C/2),
tanC=2*(1/4)/[1-(1/4)^2]=8/15.