" 5."sin(B-C)/(2)=(b-c)/(a)cos(A)/(2)

If : A+B+C= pi "then" : 1 - sin^(2)""(A)/(2) - sin^(2)""(B)/(2)+ sin^(2)""(C)/(2)= A) 2cos""(A)/(2) * cos sin ^(2)""(B)/(2) + sin^(2)""(C)/(2) B) 2 cos ""(B)/(2)* cos ""(B)/(2) * sin""(C)/(2) C) 2 cos ""(C)/(2)* cos ""(A)/(2) * sin""(B)/(2) D) 2 cos ""(A)/(2)* cos ""(B)/(2) * sin""(C)/(2)

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

If A+B+C=pi then prove that cos A+cos B+cos C=1+4sin((A)/(2))*sin((B)/(2))*sin((C)/(2))

If A , B , C are angles in a triangle , then prove that sin A + sin B + sin C =4 cos. (A)/(2) cos . (B)/(2) cos .(C)/(2)

Theorem 4:sin A+sin B+sin C=4(cos A)/(2)(cos B)/(2)(cos C)/(2)

In a triangle ABC if (sin2A+sin2B+sin2C)/(cos A+cos B+cos C-1)=((lambda)/(2))cos((A)/(2))cos((B)/(2))cos((C)/(2)) then lambda equals

(cos ^(2) ((B-C)/( 2)) )/( (b+c)^(2))+( sin ^(2)((B-C)/( 2)) )/( (b-c)^(2))=

Assertion A:In/_ABC,sum(cos A)/(sin B sin C)=2 Reason R:In/_ABC,sin A+sin B+sin C=4(cos A)/(2)(cos B)/(2)(cos C)/(2)