in triangle ABC, prve that (b+c-a) tan A/2=(c+a-b)tanB/2=(a+b-c)tan C/2 Share with your friends Share 0 Anuradha Sharma answered this So basically we have, s-atanA2=s-btanB2=s-ctanC2Putting the values of s = a+b+c2 a+b+c2-atanA2=a+b+c2-btanB2=a+b+c2-ctanC2b+c-atanA2=a+c-btanB2=a+b-ctanC2 As 12 will get cancelled Regards 8 View Full Answer