# If

Question:

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 $x+y+z=$ ____________ .

Solution:

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

$\therefore x+y+z=0$