In a `DeltaABC`, prove that `tan((A+B)/2)=cot(c/2)`

In a Delta ABC prove that (tan(A+B))/(2)=(cot C)/(2)

In a "DeltaA B C prove that tan(("A"+"B")/2)=cot("C"/2)

In a ∆ ABC prove that tan( (A+B)/2) = cot( C/2) .

In triangle ABC prove that tan((A+B)/2)=cot. C/2

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

In triangle ABC, prove that tan(B+C)/2=cot(A/2)

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

In any DeltaABC , prove that (i) tan""(B-C)/2=(b-c)/(b+c)cot""A/2 . (ii) If angleB=90^(@) , prove that tan(A/2)=sqrt(((b-c)/(b+c))) . (iii) If angleC=90^(@) , prove that tan((A−B)/2)/tan((A+B)/2)=(a-b)/(a+b) .

In any DeltaABC , prove that : tan (A/2 + B) = (c+b)/(c-b) tan (A/2)

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