Question

If $x\cos \theta =y\cos (\theta +\frac{2\pi }{3})=z\cos (\theta +\frac{4\pi }{3})$ then the value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is equal to

• A. 1
• B. 2
• C. 0
• D. $3\cos \theta$

Answer 1

The correct option is C 0

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

$\Rightarrow \cos \theta =\frac{k}{x},\cos (\theta +\frac{2\pi }{3})=\frac{k}{y}$ and $\cos (\theta +\frac{4\pi }{3})=\frac{k}{z}$

Hence, $\frac{k}{x}+\frac{k}{y}+\frac{k}{z}=\cos \theta +\cos (\theta +\frac{2\pi }{3})+\cos (\theta +\frac{4\pi }{3})$

$=\cos \theta +\cos (\theta +\pi -\frac{\pi }{3})+\cos (\theta +\pi +\frac{\pi }{3})$

$=\cos \theta +\cos (\pi -(\frac{\pi }{3}-\theta ))+\cos (\pi +(\theta +\frac{\pi }{3}))$

$=\cos \theta -\cos (\frac{\pi }{3}-\theta )-\cos (\frac{\pi }{3}+\theta )$ ..... $[\because \cos (\pi -x)=-\cos x=\cos (\pi +x\left)\right]$

$=\cos \theta -2\cos \frac{\pi }{3}\cos \theta$ ...... [Using $\cos (A+B)+\cos (A-B)=2\cos A\cos B]$

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

$=0$

Hence, option C is correct.