In any `DeltaABC` (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)`

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)

In DeltaABC prove that tan((B-C)/(2))=(b-c)/(b+c)cot.(A)/(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((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(C)/(2)

In a DeltaABC,sum(b+c)tan""A/2tan((B-C)/2)=

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

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