Question

If $x\cos \theta =y\cos (\theta +\frac{2\pi }{3})=z\cos (\theta +\frac{4\pi }{3})$, then $xy+yz+zx=0.$

• A. True
• B. False

Answer 1

The correct option is A True

$xy+yz+zx=xyz(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$

Put $x\cos \theta =y\cos (\theta +\frac{2\pi }{3})=z\cos (\theta +\frac{4\pi }{3})=k$(say)

Then

$x=\frac{k}{\cos \theta },y=\frac{k}{\cos (\theta +\frac{2\pi }{3})}$ and $z=\frac{k}{\cos (\theta +\frac{4\pi }{3})}$

So that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$

$=\frac{1}{k}[\cos \theta +\cos (\theta +\frac{2\pi }{3})+\cos (\theta +\frac{4\pi }{3})]$

$=\frac{1}{k}[\cos \theta +\cos (\theta +\pi -\frac{\pi }{3})+\cos (\theta +\pi +\frac{\pi }{3})]$

$=\frac{1}{k}[\cos \theta -\cos (\theta -\frac{\pi }{3})-\cos (\theta +\frac{\pi }{3})]$

$=\frac{1}{k}[\cos \theta -(\cos \theta \cos \frac{\pi }{3}+\sin \theta \sin \frac{\pi }{3})-(\cos \theta \cos \frac{\pi }{3}-\sin \theta \sin \frac{\pi }{3})]$

$=\frac{1}{k}[\cos \theta -\cos \theta \cos \frac{\pi }{3}-\sin \theta \sin \frac{\pi }{3}-\cos \theta \cos \frac{\pi }{3}+\sin \theta \sin \frac{\pi }{3}]$

$=\frac{1}{k}[\cos \theta -2\cos \theta \cos \frac{\pi }{3}]$

$=\frac{1}{k}\cos \theta (1-2\cos \frac{\pi }{3})$

$=\frac{1}{k}\cos \theta (1-2\times \frac{1}{2})$

$=\frac{1}{k}\cos \theta (1-1)$

$=\frac{1}{k}\cos \theta \times 0=0$

Hence the above statement is true.