" 3."(a+b)/(c)=(cos((A-B)/(2)))/(sin(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

If a,b,c are sides opposte to the angles A,B , C then which of the following is correct (1)(b+c)cos((A)/(2))=a sin((B+C)/(2))(2)(b+c)cos((B+C)/(2))=a sin((A)/(2))(3)(b-c)cos((B-C)/(2))=a(cos A)/(2)(4)(b-c)cos((A)/(2))=a sin((B-C)/(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=pi then prove that cos A+cos B+cos C=1+4sin((A)/(2))*sin((B)/(2))*sin((C)/(2))

cos A + cos B + cos C = 1 + 4sin ((A) / (2)) sin ((B) / (2)) sin ((C) / (2))

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

Theorem 3:cos A+cos B+cos C=1+4(sin A)/(2)(sin B)/(2)(sin C)/(2)

(sin2A+sin2B+sin2C)/(cos A+cos B+cos C-1)=((lambda)/(2))(cos A)/(2)(cos B)/(2)(cos C)/(2) then lambda=