" (i) "cos(A+B)/(2)=sin(C)/(2)

In a ∆ABC prove that cos((A+B)/(2))=sin(C)/(2)

If A, B , C are angles of a triangle, then P. T sin ^(2) . (A)/(2)+ sin^(2). (B)/(2) - sin ^(2). (C)/(2) =1-2 cos. (A)/(2) cos. (B)/(2) sin .(C)/(2)

If A, B , C =pi, then prove that cos ^(2). (A)/(2)+ cos^(2). (C)/(2)- cos^(2). (C)/(2)=2(cos. (A)/(2) cos. (B)/(2)sin. (C)/(2))

For DeltaABC , prove that, (a+b)/(c)=(cos((A-B)/(2)))/(sin""(C)/(2)) .

In a Delta ABC prove that cos((A+B)/(2))])=(sin C)/(2)

If A + B + C = 180^(@) , prove that cos A + cos B - cos C = -1 + 4 cos (A)/(2) cos"" (B)/(2) sin"" (C )/(2)

If A,B,C are the angles of a triangle then prove that cos A+cos B-cos C=-1+4cos((A)/(2))cos((B)/(2))sin((C)/(2))

For any triangle ABC ,prove that (a+b)/(c)=(cos((A-B)/(2)))/(sin(C)/(2))

For any triangle ABC, prove that : (a+b)/(c)=(cos((A-B)/(2)))/(sin""(C)/(2))

If A+B+C=pi then prove that cos^(2)((A)/(2))+cos^(2)((B)/(2))-cos^(2)((C)/(2))=2cos((A)/(2))cos((B)/(2))sin((C)/(2))