Prove that `(cotA/2+cotB/2)(asin^2B/2+bsin^2A/2)=ccotC/2`

In A B C ,(cot A/2+cotB/2)(asin^2B/2+bsin^2A/2)= cotC (b) ccotC (c) cotC/2 (d) ccotC/2

Prove that: cotA/2-tanA/2=(cosA/2)/(sinA/2)-(sinA/2)/(cosA/2) =(cos^(2)A/2-sin^(2)A/2)/(sinA/2cosA/2) =cosA/(1/2.(2sinA/2cosA/2)) =(2cosA)/(sinA)=2cotA = RHS. Hence Proved.

Prove that: cotA-cot2A="cosec"2A

Prove that (sin^Acos^B-cos^2Asin^2B)=(sin^2A-sin^2B)

Prove that: tanA/2+ cotA/2=2 "cosec"A

In DeltaABC , prove that: 2[asin^(2)B/2+bsin^(2)A/2]=a+b-c

Prove that : sin^(2)Acos^(2)B-cos^(2)Asin^(2)B=sin^(2)A-sin^(2)B

If A+B+C=pi , prove that : (cotA+cotB)/(tanA+tanB) + (cotB+cotC)/(tanB+tanC) + (cotC+cotA)/(tanC+tanA) =1