" (ii) "cos A sin(B-C)+sin B sin(C-A)+cos C sin(A-B)=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

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

(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,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)=

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

if A + B + C = pi then (cos A) / (sin B sin C) + (cos B) / (sin C sin A) + (cos C) / (sin A sin B) =