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.
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