`sin(B-C)cos A+sin(C-A)cos B+sin(A-B)cos C=0`

Show that: sin(B-C)cos(A-D)+sin(C-A)cos(B-D)+sin(A-B)cos(C-D)

Prove that: sin(B-C)cos(A-D)+sin(C-A)cos(B-D)+sin(A-B)cos(C-D)

If : x=sin(B-C) * cos A, y=sin(C-A) * cos B, z=sin(A-B) * cos C, then : x+y+z= (a) sin(A-B-C) (b) cos(A-B-C) (c)0 (d)-1

cos(A-D)sin(B-C)+cos(B-D)sin(C-A)+cos(C-D)sin(A-B)=0

What is the value of sin (B - C) cos (A- D) + sin (A-B) cos (C -D) + sin (C- A) cos (B - D) ? sin (B - C) cos (A- D) + sin (A-B) cos (C -D) + sin (C- A) cos (B - D) का मान क्या है?

(2sin (AC) cos C-sin (A-2C)) / (2sin (BC) cos C-sin (B-2C)) = (sin A) / (sin B)

(2sin (AC) cos C-sin (A-2C)) / (2sin (BC) cos C-sin (B-2C)) = (sin A) / (sin B)

If A,B, C are the angles of a triangle ABC then (cos A+i sin A)(cos B+i sin B)(cos C+i sin C)

Prove that (sin(B-C))/(cos B cos C)+(sin(C-A))/(cos C cos A)+(sin(A-B))/(cos A cos B)=0

Prove that: (sin(B-C))/(cos B cos C)+(sin(C-A))/(cos C cos A)+(sin(A-B))/(cos A cos B)=0