`cot(A/2)-cot(B/2)=(b-a)/Delta(M), M=`

With usual notations prove that in a triangle ABC, cot(A/2)+cot(B/2)+cot(C/2)=s^2/Delta

For any triangle ABC prove that (b-c)cot(A/2)+(c-a)cot(B/2)+(a-b)cot(C/2) =0

In a triangle ABC if cot(A/2)cot(B/2)=c, cot(B/2)cot(C/2)=a and cot(C/2)cot(A/2)=b then 1/(s-a)+1/(s-b)+1/(s-c)=

In a triangle ABC if cot(A/2)cot(B/2)=c, cot(B/2)cot(C/2)=a and cot(C/2)cot(A/2)=b then 1/(s-a)+1/(s-b)+1/(s-c)=

In any triangle ABC, prove that: (b-c)cot(A)/(2)+(c-a)cot(B)/(2)+(a-b)cot(C)/(2)=0

Prove that cot.(A)/(2)+cot.(B)/(2)+cot.(C)/(2)=(s^(2))/(Delta)

In Delta ABC the value of (cot((A)/(2))cot((B)/(2))-1)/(cot((A)/(2))cot((B)/(2))) is

In Delta ABC prove that (b-c)cot(A/2) + (c-a)cot(B/2) + (a-b)cot(C/2) = 0

In any triangle A B C , prove that: (b-c)cot (A/2)+(c-a)cot (B/2)+(a-b)cot (C/2) =0