`cot (A+B)/2.tan(A-B)/2=` (a) `(a+b)/(a-b)` (b) `(a-b)/(a+b)`

(cot (A + B)) / (2) * (tan (AB)) / (2) = (a) (a + b) / (ab) (b)

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 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 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)

cot B=2tan(A-B)rArr2tan B+cot B=

If (x)/(y)=(cos A)/(cos B)then(x tan A+y tan B)/(x+y)=cot(A+B)/(2)(b)cot(A-B)/(2)tan(A-B)/(2) (d) tan(A+B)/(2)

In DeltaABC,cot.(A+B)/(2).tan.(A-B)/(2) =

In DeltaABC,cot.(A+B)/(2).tan.(A-B)/(2) =