`(sin (A + 3 B)+ sin (3 A + B))/(sin2A + sin 2B) = 2 cos (A + B)`

(sin(A+3B)+sin(3A+B))/(sin2A+sin2B)=2cos(A+B)

(sin(A+3B)+sin(3A+B))/(sin2A+sin2B)=2cos(A+B)

sin 2A + sin 2B + sin 2 (A-B)= A) 4 sin A * sin B * sin (A-B) B) 4 sin A * cos B * cos (A-B) C) 4 cos A * sin B * cos (A-B) D) 4 cos A * cos B * sin (A-B)

(sin ^(2) A - sin ^(2) B)/( sin A cos A - sin B cos B) is equal to

sin ^ (2) A + sin ^ (2) B-sin ^ (2) C = 2sin A sin B cos C

If A and B are complementary angles, prove that : (sin A + sin B)/ (sin A - sin B) + (cos B - cos A)/ (cos B + cos A) = (2)/(2 sin^(2) A - 1)

If det [[cos (A + B), - sin (A + B), cos2Bsin A, cos A, sin B-cos A, sin A, cos B]] = 0

If det [[cos (A + B), - sin (A + B), cos2Bsin A, cos A, sin B-cos A, sin A, cos B]] = 0

If : A+B+C=pi, "then"" "sin ^(2) A +sin^(2)B - sin ^(2)C= A) 2 cos A * cos B * sin C B) 2 cos B * cos C * sin A C) 2 sin A * sin B * cos C D) 2 sin B * sin C * cos A