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Question

Prove that:
(i) cos (A+B+C)+cos (-A+B+C)+cos (A-B+C)+cos (A+B-C)sin (A+B+C)+sin (-A+B+C)+sin (A-B+C)-sin (A+B-C)=cot C

(ii) sin (B − C) cos (A − D) + sin (C − A) cos (B − D) + sin (A − B) cos (C − D) = 0

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Solution

(i)Consider LHS: cos(A+B+C) + cos(-A+B+C) + cos(A-B+C) + cos(A+B-C)sin(A+B+C) + sin(-A+B+C) + sin(A-B+C) - sin(A+B-C)=2cosA+B+C-A+B+C2cosA+B+C+A-B-C2+2cosA-B+C+A+B-C2cosA-B+C-A-B+C22sinA+B+C-A+B+C2cosA+B+C+A-B-C2+2sinA-B+C-A-B+C2cosA-B+C+A+B-C2=2cos B+C cos A +2cos A cos -B+C2sin B+C cos A+ 2sin -B+C cos A=2cos Acos B+C + cos-B+C2cos AsinB+C + sin-B+C=cos B+C + cos -B+CsinB+C + sin -B+C=2cos B+C-B+C2 cos B+C+B-C22sinB+C-B+C2 cos B+C+B-C2= cos C cos Bsin C cos B= cotC= RHSHence, LHS = RHS.

(ii)Consider LHS:sin (B-C) cos (A-D) + sin (C-A) cos (B-D) + sin (A-B) cos (C-D)Multiplying by 2: 2sin (B-C) cos (A-D) + 2sin(C-A) cos (B-D) + 2sin (A-B) cos (C-D)= sin B-C+A-D + sin B-C-A+D + sin C-A+B-D + sin C-A-B+D + sin A-B+C-D + sin A-B-C+D= sin-C+D-A-B+sin-A+C-B-D+sin-A+D-B-C+sinC-A-B+D+sinA-B+C-D+sinA-B-C+D=-sinC+D-A-B-sinA+C-B-D-sinA+D-B-C+sinC-A-B+D+sinA-B+C-D+sinA-B-C+D=0= RHSHence, LHS=RHS.

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