Show that: `sin(B-C)cos(A-D)+sin(C-A)cos(B-D)+sin(A-B)cos(C-D)=0`

Prove 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) = 0

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

Prove the following : cos(A-D)sin(B-C)+cos(B-D)sin(C-A)+cos(C-D)sin(A-B) = 0

If A,B,C,D are four angles then sin (A-B) cos (A+B) + sin B (B-C) .cos (B+C) + sin (C-D) cos (C+D)+ sin (D-A) cos (D+A)=

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?

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

If A, B, C, D are four angles then sin (AB) .cos (A + B) + sin (BC) .cos (B + C) + sin (CD) .cos (C + D) + sin ( DA) .cos (D + A) =

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