Question

If $x\cos \alpha =y\cos \left(2\pi /3+\alpha \right)=\cos \left(4\pi /3+\alpha \right),$ then $xy+zy+zx=$

• A. $0$
• B. $1$
• C. $-1$$
• D. $2$

Answer 1

Answer: (C)

$-1$$

Given

$x\cos \alpha =y\cos (\frac{2\pi }{3}+\alpha )=z\cos (\frac{4\pi }{3}+\alpha )$ …………$\left(1\right)$

Let $\left(1\right)$ $\to$ K(constant)

Then

$x=\frac{k}{\cos \theta }$

$y=\frac{k}{\cos 120+\theta }$

$z=\frac{k}{\cos (240+\theta )}$

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

$\frac{\cos \theta }{k}+\frac{\cos 120+\theta }{k}+\frac{\cos 240+\theta }{k}=\frac{xy+yz+zx}{xyz}$

$\frac{\cos \theta +\cos (120+\theta )+\cos (240+\theta )}{k}=\frac{xy+yz+zx}{xyz}$

$\frac{\cos \theta +2\left(\cos \frac{(120+\theta +240+\theta )}{2}\cos \frac{(120+\theta -240-\theta )}{2}\right)}{k}=\frac{xy+yz+zx}{xyz}$

$\frac{\cos {\theta }^{-}+2\left(\cos \right(180+\theta \left)\cos \right(120\left)\right)}{k}$

$\frac{\cos \theta +(-\cos \theta +\frac{1}{2})}{k}$ $=\frac{\cos \theta -\cos \theta }{k}=\frac{xyyz+zx}{xyz}$

$xy+yz+zx=0$.