Question

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

• A. $-1$
• B. $1$
• C. $0$
• D. $2$

Answer 1

Answer: (C)

$0$

Let $x\cos \theta =y\cos (\theta +\frac{2\pi }{3})=z\cos (\theta +\frac{4\pi }{3})=\lambda$

Therefore,

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

Hence

$xy+yz+zx$

$=\frac{\lambda }{\cos \theta }\cdot \frac{\lambda }{\cos (\theta +\frac{2\pi }{3})}+\frac{\lambda }{\cos (\theta +\frac{2\pi }{3})}\cdot \frac{\lambda }{\cos (\theta +\frac{4\pi }{3})}+\frac{\lambda }{\cos (\theta +\frac{4\pi }{3})}\cdot \frac{\lambda }{\cos \theta }$

$={\lambda }^2\left(\frac{\cos (\theta +\frac{4\pi }{3})+\cos \theta +\cos (\theta +\frac{2\pi }{3})}{\cos (\theta +\frac{4\pi }{3})\cdot \cos \theta \cdot \cos (\theta +\frac{2\pi }{3})}\right)$

$={\lambda }^2\left(\frac{\cos (\pi +(\frac{\pi }{3}+\theta ))+\cos \theta +\cos (\pi -(\frac{\pi }{3}-\theta ))}{\cos (\theta +\frac{4\pi }{3})\cdot \cos \theta \cdot \cos (\theta +\frac{2\pi }{3})}\right)$

$={\lambda }^2\left(\frac{-\cos (\frac{\pi }{3}+\theta )+\cos \theta -\cos (\frac{\pi }{3}-\theta )}{\cos (\theta +\frac{4\pi }{3})\cdot \cos \theta \cdot \cos (\theta +\frac{2\pi }{3})}\right)$

$={\lambda }^2\left(\frac{\cos \theta -\left(2\cos \frac{\pi }{3}\cos \theta \right)}{\cos (\theta +\frac{4\pi }{3})\cdot \cos \theta \cdot \cos (\theta +\frac{2\pi }{3})}\right)$

$={\lambda }^2\left(\frac{\cos \theta -\left(\cos \theta \right)}{\cos (\theta +\frac{4\pi }{3})\cdot \cos \theta \cdot \cos (\theta +\frac{2\pi }{3})}\right)=0$