[tan(B-C)/(2)=(b-c)/(b+c)cot(A)/(2)],[tan(C-A)/(2)=(c-a)/(c+a)cot(B)/(2)],[tan(A-B)/(2)=(a-b)/(a+b)cot(C)/(2)]

In triangle ABC,, prove that (tan(B-C))/(2),=[(b-c)/(b+c)](cot A)/(2)(tan(C-A))/(2),=[(c-a)/(c+a)](cot B)/(2)(tan(A-B))/(2),=[(a-b)/(a+b)](cot B)/(2)

In any Delta ABC(i)tan((B-C)/(2))=((b-c)/(b+c))(cot A)/(2)(ii)tan((A-B)/(2))=((a-b)/(a+b))(cot C)/(2)(iii)tan((C-A)/(2))=((c-a)/(c+a))(cot B)/(2)

Prove that tan((A-B)/2)= (a-b)/(a+b)cot frac(c)(2)

tan((A+B)/(2))="cot"(C)/(2)

cot((A)/(2))+cot((B)/(2))+cot((C)/(2))=cot((A)/(2))cot((B)/(2))cot((C)/(2))

If A+B+C=pi, show that : tan.(A)/(2)tan.(B)/(2)+tan.(B)/(2)tan.(C)/(2)+tan.(C)/(2)tan.(A)/(2)=1 Hence deduce that : cot.(A)/(2)+cot.(B)/(2)+.cot.(C)/(2)=cot.(A)/(2).cot.(B)/(2)tan.(C)/(2) .

Delta ABC;(b-c)cot((A)/(2))+(c-a)cot((B)/(2))+(a-b)cot((C)/(2))

Delta ABC;(b-c)(cot A)/(2)+(c-a)(cot B)/(2)+(a-b)(cot C)/(2)