If $x \cos \theta=y \cos \left(\theta+\frac{2 \pi}{3}\right)=z \cos \left(\theta+\frac{4 \pi}{3}\right)$, then write the value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$
Given:
$x \cos \theta=y\left(\cos \theta \cos \frac{2 \pi}{3}-\sin \theta \sin \frac{2 \pi}{3}\right)=z\left(\cos \theta \cos \frac{4 \pi}{3}-\sin \theta \sin \frac{4 \pi}{3}\right)$
$\Rightarrow x \cos \theta=y\left(-\frac{1}{2} \cos \theta-\frac{\sqrt{3}}{2} \sin \theta\right)=z\left(-\frac{1}{2} \cos \theta+\frac{\sqrt{3}}{2} \sin \theta\right)$
$\Rightarrow x=\frac{y}{2}(-1-\sqrt{3} \tan \theta)=\frac{z}{2}(-1+\sqrt{3} \tan \theta)$
$x=\frac{y}{2}(-1-\sqrt{3} \tan \theta)$
$z=\frac{y(-1-\sqrt{3} \tan \theta)}{(-1+\sqrt{3} \tan \theta)}$
Now,
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{2}{y(-1-\sqrt{3} \tan \theta)}+\frac{1}{y}+\frac{(-1+\sqrt{3} \tan \theta)}{y(-1-\sqrt{3} \tan \theta)}$
$=\frac{2+(-1-\sqrt{3} \tan \theta)+(-1+\sqrt{3} \tan \theta)}{y(-1-\sqrt{3} \tan \theta)}$
= 0