Question

Prove that:a(sinB-sinC)+b(sinC-sinA)+c(sinA-sinB)=0

Answer 1

Answer:

Step-by-step explanation:

From Laws of sine, we have,

$\tt{\blue{\dfrac{sin(A)}{a}=\dfrac{sin(B)}{b}=\dfrac{sin(C)}{c}}}$

Let

$\sf{\dfrac{sin(A)}{a}=\dfrac{sin(B)}{b}=\dfrac{sin(C)}{c}=k}$

$\sf{\implies\,sin(A)=ak;\,\,\,\,sin(B)=bk;\,\,\,\,sin(C)=ck}$

Now,

$\sf{a(sinB-sinC)+b(sinC-sinA)+c(sinA-sinB)}$

$\sf{=a(bk-ck)+b(ck-ak)+c(ak-bk)}$

$\sf{=abk-cck+bck-abk+ack-bck}$

$\sf{=0}$