Question

In ΔABC, prove that (b - a cos C) sin A = a cos A sin C.

Answer 1

By Law of Sines, we have

a / sin A  = b / sin B

b / sin B  = a / sin A

b sin A  = a sin B                 By Allied Angles   sin B = sin (π - B)

so we have

b sin A  = a sin (π - B)

            ∵ in any triangle   A + B + C = π   ⇒  A + C = π - B

b sin A  = a sin (A + C)

b sin A  = a (sin A cos C  + cos A sin C )

b sin A  = a (cos A sin C + cos C sin A)

b sin A  = a cos A sin C + a cos C sin A

b sin A - a cos C sin A = a cos A sin C

⇒  (b - a cos C) sin A = a cos A sin C

Which is the Required.

Answer 2

Answer:

this is the answer in this question