证明三角恒等式

证明:
三角形中
(tanA/2)^2+(tanB/2)^2+(tanC/2)^2=1
恒成立...

应该学过余弦定理吧
先化简下面这个式子
tan(a/2)=sin(a/2)/cos(a/2)
=[2sin(a/2)cos(a/2)]/[2cos²(a/2)]
=sina/(1+cosa)
故tan²(a/2)=sin²a/(1+cosa)²
=(1-cos²a)/(1+cosa)²
=(1-cosa)/(1+cosa)
故
(tanA/2)^2=(1-cosA)/(1+cosA)
=[1-(b²+c²-a²)/(2bc)]/[1+(b²+c²-a²)/(2bc)]
=[2bc-(b²+c²-a²)]/[2bc+(b²+c²-a²)]
=(2bc-b²-c²+a²)/(b²+c²+2bc-a²)
=[a²-(b-c)²]/[(b+c)²-a²]
=[(a+b-c)*(a-b+c)]/[(a+b+c)*(b+c-a)]
同理，有
(tanB/2)^2=[(a+b-c)*(-a+b+c)]/[(a+b+c)*(a+c-b)]
(tanC/2)^2=[(-a+b+c)*(a-b+c)]/[(a+b+c)*(b+a-c)]
故(tanA/2)^2+(tanB/2)^2+(tanC/2)^2
=[(a+b-c)*(a-b+c)]/[(a+b+c)*(b+c-a)]+[(a+b-c)*(-a+b+c)]/[(a+b+c)*(a+c-b)]+[(-a+b+c)*(a-b+c)]/[(a+b+c)*(b+a-c)]
=1

应该补上条件：三角形abc中， （a²-b²)/(cosa+cosb）+（b²-c²)/(cosb+cosc）+（c²-a²)/(cosc+cosa）=0 证明： 利用正弦定理a/(sina)=b/(sinb)=c/(sinc)=2r, 就有： a²=4r²sin^2a b²=4r²sin^2b c²=4r²sin^2c (a²-b²)=4r²(sin²a-sin²b) =4r²(1-cos²a-1+cos²b) =4r²(cos²b-cos²a) =4r²(cosa+cosb)(cosb-cosa) -------------（1）式 同理，可得 (b²-c²)=4r²(sin²b-sin²c) =4r²(cosb+cosc)(cosc-cosb) -------------（2）式 (c²-a²)=4r²(sin²c-sin²a) =4r²(cosc+cosa)(cosa-cosc）…………（3）式 (a²-b²)/(cosa+cosb)+(b²-c²)/(cosb+cosc)+(c²-a²)/(cosc+cosa) =4r²(cosb-cosa)+4r²(cosc-cosb)+4r²(cosa-cosc) =4r²［(cosb-cosa)+(cosc-cosb)+(cosa-cosc)］ =0 完毕