Question

If $x\sin \theta =y\sin (\theta +\frac{2\pi }{3})=z\sin (\theta +\frac{4\pi }{3})$, then

• A. $x+y+z=0$
• B. $xy+yz+zx=0$
• C. $xyz+x+y+z=1$
• D. None of the above

Answer 1

Answer: (B)

$xy+yz+zx=0$

We have,
$x\sin \theta =y\sin (\theta +\frac{2\pi }{3})=z\sin (\theta +\frac{4\pi }{3})$
$\Rightarrow \frac{\sin \theta }{1/x}=\frac{\sin (\theta +\frac{2\pi }{3})}{1/y}=\frac{\sin (\theta +\frac{4\pi }{3})}{1/z}$
$\Rightarrow \frac{\sin \theta }{1/x}=\frac{\sin \left(\theta +\frac{2\pi }{3}\right)}{1/y}=\frac{\sin \left(\theta +\frac{4\pi }{3}\right)}{1/z}$
$=\frac{\sin \theta +\sin (\theta +2\pi /3)+\sin (\theta +4\pi /3)}{1/x+1/y+1/z}$
$\Rightarrow \frac{\sin \theta }{1/x}=\frac{\sin (\theta +2\pi /3)}{1/y}=\frac{\sin (\theta +4\pi /3)}{1/z}$
$=\frac{\sin \theta +2\sin (\pi +\theta )\cos \pi /3}{1/x+1/y+1/z}$
$\Rightarrow \frac{\sin \theta }{1/x}\times \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=0$
$\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$
$\Rightarrow xy+yz+zx=0$