`A+B+C=18 0^(@)=>C o s e c A[Sin B cos C+ cos B sin c]=`

If A+B+C=180^0,\ then s e c A(cos B cos C-sin B sin C) is equal to:

If A+B+c=180^(@), then sec A(cos B cos C-sin B sin C) is equal to 0 b.-1 c.1d .none of these

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

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

If : A+B+C=pi, "then" : 1-cos^(2) A-cos^(2)B-cos^(2)C= A) 2 cos A * cos B * cos C B) 2 sin A * cos B * cos C C) 2 cos A * sin B * cos C D) 2 cos A * cos B * sin C

Expand cos ( A + B + C). Hence prove that cos A cos B cos C = sin A sin B cos C + sin B sin C cos A + sin C sin A cos B, if A + B + C = (pi)/(2) .

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

If A+B+C=180^(@)C then (sin 2A+sin 2B+sin 2C)/(cos A +cos B +cos C-1)=

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