Question

If $x\cos \theta =y\cos \left(\theta +\frac{2\pi }{3}\right)=z\cos \left(\theta +\frac{4\pi }{3}\right)$, 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

$0$

Explanation for the correct option.

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

Now,

$$\begin{gathered} \frac{k}{x}=\cos \theta .....\left(1\right) \\ \frac{k}{y}=\cos \left(\theta +\frac{2\mathrm{\pi }}{3}\right)......\left(2\right) \\ \frac{k}{z}=\cos \left(\theta +\frac{4\mathrm{\pi }}{3}\right)......\left(3\right) \end{gathered}$$

By adding $\left(1\right),\left(2\right),\mathrm{and}\left(3\right)$ we get

$$\begin{gathered} \frac{k}{x}+\frac{k}{y}+\frac{k}{z}=\cos \theta +\cos \left(\theta +\frac{2\mathrm{\pi }}{3}\right)+\cos \left(\theta +\frac{4\mathrm{\pi }}{3}\right) \\ \Rightarrow k\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\cos \theta +\cos \left(\theta +\frac{2\mathrm{\pi }}{3}\right)+\cos \left(\theta +\frac{4\mathrm{\pi }}{3}\right) \\ =2\cos \left(\frac{2\theta +{\displaystyle \frac{4\mathrm{\pi }}{3}}}{2}\right)\cos \left(\frac{-{\displaystyle \frac{4\mathrm{\pi }}{3}}}{2}\right)+\cos \left(\theta +\frac{2\mathrm{\pi }}{3}\right)\left[\because cosA+cosB=2cos\left(\frac{A+B}{2}\right)cos\left(\frac{A-B}{2}\right)\right] \\ =-2\cos \left(\theta +\frac{2\mathrm{\pi }}{3}\right)\cos \left(\frac{2\mathrm{\pi }}{3}\right)+\cos \left(\theta +\frac{2\mathrm{\pi }}{3}\right) \\ =-2\times \frac{1}{2}\cos \left(\theta +\frac{2\mathrm{\pi }}{3}\right)+\cos \left(\theta +\frac{2\mathrm{\pi }}{3}\right)\left[by\cos \left(\frac{2\mathrm{\pi }}{3}\right)=\frac{1}{2}\right] \\ =0 \end{gathered}$$

Hence, option C is correct.