If $\frac{x}{\cos \theta}=\frac{y}{\cos \left(\theta-\frac{2 \pi}{3}\right)}=\frac{z}{\cos \left(\theta+\frac{2 \pi}{3}\right)}$, then $x+y+z=$ ____________ .
$\frac{x}{\cos \theta}=\frac{y}{\cos \left(\theta-\frac{2 \pi}{3}\right)}=\frac{z}{\cos \left(\theta+\frac{2 \pi}{3}\right)}$ say $=K$
i. e. $x=K \cos \theta, y=K \cos \left(\theta-\frac{2 \pi}{3}\right), z=K \cos \left(\theta+\frac{2 \pi}{3}\right)$
$\therefore x+y+z=K\left[\cos \theta+\cos \left(\theta-\frac{2 \pi}{3}\right)+\cos \left(\theta+\frac{2 \pi}{3}\right)\right]$
[using identity : $\left.\cos a+\cos b=2 \cos \left(\frac{a+b}{2}\right) \cos \left(\frac{a-b}{2}\right)\right]$
$=K[\cos \theta+2 \cos \theta \cos (2 \pi / 3)]$ $\left(\because \frac{\theta-\frac{2 \pi}{3}+\theta+\frac{2 \pi}{3}}{2}=0\right.$ and $\left.\cos (-x)=\cos x\right)$
$=K\left[\cos \theta+2 \cos \theta\left(-\frac{1}{2}\right)\right]$ $\left[\because \cos \frac{2 \pi}{3}=-\frac{1}{2}\right]$
$=k[\cos \theta-\cos \theta]$
= 0
$\therefore x+y+z=0$
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