In triangle ABC, prove the following:
Question:

In triangle ABC, prove the following:

$a(\sin B-\sin C)+(\sin C-\sin A)+c(\sin A-\sin B)=0$

Solution:

Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

Then,

Consider the LHS of the equation $a(\sin B-\sin C)+(\sin C-\sin A)+c(\sin A-\sin B)=0$.

LHS $=a(\sin B-\sin C)+b(\sin C-\sin A)+c(\sin A-\sin B)$

$=k \sin A(\sin B-\sin C)+k \sin B(\sin C-\sin A)+k \sin C(\sin A-\sin B)$

$=k \sin A \sin B-k \sin A \sin C+k \sin B \sin C-k \sin B \sin A+k \sin C \sin A-k \sin C \sin B$

$=0=$ RHS

Hence proved.