Question

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

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is B $0$

We have $x=y\cos \frac{2\pi }{3}=z\cos \frac{4\pi }{3}$ (say)

$\Rightarrow \frac{1}{x}=\frac{1}{k},\frac{1}{y}=\frac{\cos \frac{2\pi }{3}}{h},\frac{1}{z}=\frac{\cos \frac{4\pi }{3}}{k}$

$\therefore \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{k}(1+\cos \frac{2\pi }{3}+\cos \frac{4\pi }{3})=\frac{1}{k}(1-\frac{1}{2}-\frac{1}{2})=0$

$\Rightarrow xy+yz+xz=0$