Question

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

• A. $1$
• B. $0$
• C. $-1$
• D. none

Answer 1

The correct option is B $0$

Let $\frac{x}{\cos \theta }=\frac{y}{\cos (\theta -\frac{2\pi }{3})}=\frac{z}{\cos (\theta +\frac{2\pi }{3})}=k$ where $k$ is a constant.

$\Rightarrow \frac{x}{\cos \theta }=k,\frac{y}{\cos (\theta -\frac{2\pi }{3})}=k,\frac{z}{\cos (\theta +\frac{2\pi }{3})}=k$

$\Rightarrow x=k\cos \theta ,y=k\cos (\theta -\frac{2\pi }{3}),z=k\cos (\theta +\frac{2\pi }{3})$

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

$=k[\cos \theta +\cos (\theta -\frac{2\pi }{3})+\cos (\theta +\frac{2\pi }{3})]$

$=k[\cos \theta +2\cos \left(\frac{\theta -\frac{2\pi }{3}+\theta +\frac{2\pi }{3}}{2}\right)\cos \left(\frac{\theta -\frac{2\pi }{3}-\theta -\frac{2\pi }{3}}{2}\right)]$

$=k[\cos \theta +2\cos \left(\frac{2\theta }{2}\right)\cos \left(\frac{-2\frac{2\pi }{3}}{2}\right)]$

$=k[\cos \theta +2\cos \theta \cos (-\frac{\frac{4\pi }{3}}{2})]$

$=k[\cos \theta +2\cos \theta \cos \left(\frac{2\pi }{3}\right)]$

$=k[\cos \theta +2\cos \theta \cos (\pi -\frac{\pi }{3})]$

$=k[\cos \theta -2\cos \theta \cos \frac{\pi }{3}]$

$=k[\cos \theta -2\cos \theta \times \frac{1}{2}]$

$=k[\cos \theta -\cos \theta ]=0$