`2cotA="cot"(A)/(2)-"tan"(A)/(2)`.

cot A =1/2 (cot ""(A)/(2) - tan "" (A)/(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) .

cot A=(1)/(2)(cot((A)/(2))-tan((A)/(2)))

cot ((A)/(2))-tan((A)/(2))= A) 2 sin A B) 2 cos A C) 2 tan A D) 2 cot A

If A+B+C=pi , prove that (a) tan 3A+tan3B+tan3C=tan3A tan 3B tan 3C (b) cot""(A)/(2)+cot""(B)/(2)+cot""(C)/(2)=cot""(A)/(2)cot""(B)/(2)cot""(C)/(2)

In DeltaABC , with usual notation show that ((a+b+c)^(2))/(a^(2)+b^(2)+c^(2))=(cot.(A)/(2)+cot.(B)/(2)+cot.(C)/(2))/(cotA+cotB+cotC)

Prove the following : cotA/(cotA-cot3A)-tanA/(tan3A-tanA) =1

Prove that cosA=(cotA)/("cosec A")=(cotA)/(sqrt((1+cot^(2)A)))" "(A in" III quad.")

Prove the following identities. where the angles involved are acute angles for which the expressions are defined. ((1+tan^(2)A)/(1+cot^(2)A))=((1-tanA)/(1-cotA))^(2)=tan^(2)A