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

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

In DeltaABC,(cot(A//2)+cot(B//2)+cot(C//2))/(cotA+cotB+cotC)=

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

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) .

Prove: (1-tan^(2)A)/(cot^(2)A-1)=tan^(2)A

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

(tan A)/((1+tan^(2)A)^(2))+(cot A)/((1+cot^(2)A)^(2))=sin A cos A

Prove that cot2A=(cot^2A-1)/(2cotA)

Prove that (1+tan^(2)A)/(1+cot^(2)A)=((1-tan A)/(1-cot A))^(2)=tan^(2)A