Question

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

• A. $1$
• B. $0$
• C. $-1$
• D. None of these

Answer 1

The correct option is B

$0$

Explanation for the correct option.

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

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

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

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

Hence, option B is correct.