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

In triangle ABC,a,b,c are the lengths of its sides and A,B,C are the angles of triangle ABC .The correct relation is given by (a) (b-c)sin((B-C)/(2))=a(cos A)/(2) (b) (b-c)cos((A)/(2))=as in(B-C)/(2)(c)(b+c)sin((B+C)/(2))=a(cos A)/(2)(d)(b-c)cos((A)/(2))=2a(sin(B+C))/(2)

For triangle ABC, show that: sin((A+B)/(2))-cos(C)/(2)=0

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

If A+B+C=pi , prove that: "sin" A+"sin" B-"sin" C=4 "sin"(A)/(2)"sin"(B)/(2)"cos"(C)/(2) .

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

Which of the following is true in a triangle ABC?(1)(b+c)sin((B_(C))/(2))=2a cos((A)/(2))(2+c)cos((A)/(2))=2a sin((B-C)/(2))

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

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

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

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