`(cos A)/(sin B sin C)+(cos B)/(sin C sin A)+(cos C)/(sin A sin B)=2`

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

(cos A) / (c sin B) + (cos B) / (a sin C) + (cos C) / (b sin A) =

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

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

If : A+B+C=pi, "then" : sin 2A + sin 2 B - sin 2 C= A) 4 sin A * cos B * cos C B) 4 sin B * sin C * cos A C) 4 sin C * cos A * cos B D) 4 sin A * sin B * sin C

In any triangle ABC,sin^(2)A-sin^(2)B+sin^(2)C is always equal to (A) 2sin A sin B cos C(B)2sin A cos B sin C(C)2sin A cos B cos C(D)2sin A sin B sin C

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

Assertion A:In/_ABC,sum(cos A)/(sin B sin C)=2 Reason R:In/_ABC,sin A+sin B+sin C=4(cos A)/(2)(cos B)/(2)(cos C)/(2)

(sin (BC)) / (cos B cos C) + (sin (CA)) / (cos C cos A) + (sin (AB)) / (cos A cos B) = 0

(b-a cos c)sin A=a cos A sin c