Question

化简tanA/2tanB/2+tanB/2tanC/2+tanC/2tanA/2拜托各位大神

Answer 1

应该有条件“A、B、C为三角形内角”吧. 如果A+B+C=π,即A、B、C为三角形内角,故：A/2+B/2+C/2=π/2 则：tan(A/2+B/2)=cot(C/2)=1/ tan(C/2),tan(A/2+C/2)=cot(B/2)=1/ tan(B/2) ,tan(B/2+C/2)=cot(A/2)=1/ tan(A/2) 又：tan(A/2+B/2)=[tan(A/2)+tan(B/2)]/[1- tan(A/2) tan(B/2)] 故：tan(A/2) tan(B/2)= 1-[tan(A/2)+tan(B/2)]/ tan(A/2+B/2)=1-[tan(A/2)+tan(B/2)] tan(C/2) 同理：tan(B/2)tan(C/2)= 1-[tan(C/2)+tan(B/2)]/ tan(C/2+B/2)= 1-[tan(C/2)+tan(B/2)] tan(A/2)；tan(C/2)tan(A/2)= 1-[tan(C/2)+tan(A/2)]/ tan(C/2+A/2)=1-[tan(C/2)+tan(A/2)] tan(B/2) 故：tan(A/2) tan(B/2)+ tan(B/2)tan(C/2)+ tan(C/2)tan(A/2)= 1-[tan(A/2)+tan(B/2)] tan(C/2)+ 1-[tan(C/2)+tan(B/2)] tan(A/2)+ 1-[tan(C/2)+tan(A/2)] tan(B/2) =3-2tan(A/2) tan(B/2)-2tan(B/2)tan(C/2)-2tan(C/2)tan(A/2) 故：3tan(A/2) tan(B/2)+3tan(B/2)tan(C/2)+3tan(C/2)tan(A/2)=3 故：tan(A/2) tan(B/2)+ tan(B/2)tan(C/2)+ tan(C/2)tan(A/2)= 1