# If x cosθ=y cos

Question:

If $x \cos \theta=y \cos \left(\theta+\frac{2 \pi}{3}\right)=z \cos \left(\theta+\frac{4 \pi}{3}\right)$, prove that $x y+y z+z x=0$      [NCERT EXEMPLAR]

Solution:

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

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

$\Rightarrow \frac{\cos \theta}{\frac{1}{x}}=\frac{\cos \left(\theta+\frac{2 \pi}{3}\right)}{\frac{1}{y}}=\frac{\cos \left(\theta+\frac{4 \pi}{3}\right)}{\frac{1}{z}}=\frac{\cos \theta+\cos \left(\theta+\frac{2 \pi}{3}\right)+\cos \left(\theta+\frac{4 \pi}{3}\right)}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$$\left(\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\ldots=\frac{a+c+e+\ldots}{b+d+f+\ldots}\right) \Rightarrow \frac{\cos \theta}{\frac{1}{x}}=\frac{\cos \left(\theta+\frac{2 \pi}{3}\right)}{\frac{1}{y}}=\frac{\cos \left(\theta+\frac{4 \pi}{3}\right)}{\frac{1}{z}}=\frac{\cos \theta+2 \cos \left(\frac{\theta+\frac{2 \pi}{3}+\theta+\frac{4 \pi}{3}}{2}\right) \cos \left(\frac{\theta+\frac{2 \pi}{3}-\theta-\frac{4 \pi}{3}}{2}\right)}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \Rightarrow \frac{\cos \theta}{\frac{1}{x}}=\frac{\cos \left(\theta+\frac{2 \pi}{3}\right)}{\frac{1}{y}}=\frac{\cos \left(\theta+\frac{4 \pi}{3}\right)}{\frac{1}{z}}=\frac{\cos \theta+2 \cos (\pi+\theta) \cos \left(-\frac{\pi}{3}\right)}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \Rightarrow \frac{\cos \theta}{\frac{1}{x}}=\frac{\cos \left(\theta+\frac{2 \pi}{3}\right)}{\frac{1}{y}}=\frac{\cos \left(\theta+\frac{4 \pi}{3}\right)}{\frac{1}{z}}=\frac{\cos \theta+2 \times(-\cos \theta) \times \frac{1}{2}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$$[\cos (-\theta)=\cos \theta]$

$\Rightarrow \frac{\cos \theta}{\frac{1}{x}}=\frac{\cos \left(\theta+\frac{2 \pi}{3}\right)}{\frac{1}{y}}=\frac{\cos \left(\theta+\frac{4 \pi}{3}\right)}{\frac{1}{z}}=\frac{0}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$

$\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$

$\Rightarrow \frac{y z+z x+x y}{x y z}=0$

$\Rightarrow x y+y z+z x=0$