" i) "cot(A)/(2)+cot(B)/(2)+cot(C)/(2)=(s^(2))/(Delta)

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

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

Show that cot ""(A)/(2) + cot "" ( B)/(2) + cot "" ( C ) /(2) = (S^(2))/( Delta )

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 Delta ABC , if a=3 , b=4,c=6 then ("cot" (A)/(2)+"cot" (B)/(2)+ "cot" ( C)/(2))/("cot" A + "cot" B + "cot" C)=

In a DeltaABC , 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) equals

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

In Delta ABC, "cot"(A)/(2) + "cot" (B)/(2) + "cot" (C)/(2) is equal to