Question

cosA sin(B-c) +sinB sin(c-a) + cosc sin (a-b) =0​

Answer 1

Answer:

To Prove: cos A . sin ( B - C ) + cos B . sin ( C - A ) + cos C . sin ( A - B ) = 0

We use the the following result,

sin ( x - y ) = sin x . cos y - cos x . sin y

LHS

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

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

= cos A . sin B . cos C - cos A . cos B . sin C + cos B . sin C . cos A - cos B . cos C . sin A + cos C . sin A . cos B - cos C . cos A . sin B

= 0 + 0 + 0

= 0

RHS

= 0

LHS = RHS

Hence, Proved

Answer 2

Answer:

Answer

The proof is explained below :

Step-by-step explanation:

Using Sine rule in a triangle ,

Now, LHS = a·sinA - b·sinB

= k·sin²A - k·sin²B { Using Sine rule equations }

= k·[sin(A + B)·sin(A - B)]

= k·sin(π - C)·sin(A - B) { By angle sum property of a triangle sum of all angles is 180 ⇒ A + B + C = π ⇒ A + B = π - C }

= k·sinC·sin(A - B)

= c·sin(A - B) = R.H.S.

So, LHS = RHS

Hence, a·sinA - b·sinB = c·sin(A - B)