解析：∵tanA：tanB：tanC=1:2：3,则tanA=k,tanB=2k,tanC=3k即sinA/k=cosA,sinB/2k=cosB,sinC/3k=cosC(sinA/k)^2+(sinA)^2=1,得（sinA)^2=k^2/(k^2+1)同理(sinB)^2=4k^2/(4k^2+1)( sinC)^2=9k^2/(9k^2+1)∵tanC=-tan...