2014/086) if $a\sin\,x=b\sin(x+\frac{2\pi}{3})=c\sin(x+\frac{4\pi}{3})$

prove that $ab+bc+ca=0$ and $\dfrac{1}{a} +\dfrac{1}{b}+\dfrac{1}{c}=ab+bc+ca$

proof
say
$a\sin\,x=b\sin(x+\dfrac{2\pi}{3})=c\sin(x+\dfrac{4\pi}{3})=k$
  then
$\sin\, x= \dfrac{k}{a}$
 $\sin(x+\dfrac{2\pi}{3})=\dfrac{k}{b}$
 $\sin(x+\dfrac{4\pi}{3})=\dfrac{k}{c}$
  

so $k(\dfrac{1}{a} + \dfrac{1}{b} +  \dfrac{1}{c}) = ( \sin\,x +\sin(x+\frac{2\pi}{3})+ \sin(x+\frac{4\pi}{3}))$
=$\sin\, x + 2 \cos \dfrac{2\pi}{3} \sin\, x$
=  $\sin\, x + 2 \cdot \dfrac{-1}{2} \sin\, x$
=  0

as k is not zero

$\dfrac{1}{a} + \dfrac{1}{b} +  \dfrac{1}{c}= 0$  and multiply both sides by abc to get

bc + ca + ab = 0

as both are zero they are same