Question

If $x\sin \theta =y\sin (\theta +2\pi /3)=z\sin (\theta +4\pi /3)$ then $xy+yz+zx$ is equal to:

• A. $1$
• B. $\frac{1}{2}$
• C. $0$
• D. None of these

Answer 1

The correct option is C

$0$

Explanation for the correct option.

Step 1: Simplify $x\sin \theta =y\sin (\theta +2\pi /3)$

$\begin{array}{rcl}x\sin \theta & =& y\sin \left(\theta +\frac{2\pi }{3}\right)\\ & \Rightarrow & x\sin \theta =y\left(\sin \theta \cos \frac{2\pi }{3}+\sin \frac{2\pi }{3}\cos \theta \right)\left[by\sin \left(a+b\right)=\sin a\cos b+\sin b\cos a\right]\\ & \Rightarrow & x\sin \theta =y\left[\sin \theta \left(-\frac{1}{2}\right)+\left(\frac{\sqrt{3}}{2}\right)\cos \theta \right]\\ \Rightarrow \frac{x}{y}& =& \frac{-{\displaystyle \frac{1}{2}}\sin \theta +\frac{\sqrt{3}}{2}\cos \theta }{\sin \theta }\\ & =& -\frac{1}{2}+\frac{\sqrt{3}}{2}\cot \theta ...\left(1\right)\end{array}$

Step 2: Simplify $x\sin \theta =z\sin (\theta +4\pi /3)$

$\begin{array}{rcl}x\sin \theta & =& z\sin \left(\theta +\frac{4\pi }{3}\right)\\ & \Rightarrow & x\sin \theta =z\left(\sin \theta \cos \frac{4\pi }{3}+\sin \frac{4\pi }{3}\cos \theta \right)\\ & \Rightarrow & x\sin \theta =z\left[\sin \theta \left(-\frac{1}{2}\right)-\left(\frac{\sqrt{3}}{2}\right)\cos \theta \right]\\ \Rightarrow \frac{x}{z}& =& \frac{-\frac{1}{2}\sin \theta -\frac{\sqrt{3}}{2}\cos \theta }{\sin \theta }\\ & =& -\frac{1}{2}-\frac{\sqrt{3}}{2}\cot \theta ......\left(2\right)\end{array}$

Step 3: Add $\left(1\right)\mathrm{and}\left(2\right)$

$\begin{array}{rcl}\frac{x}{y}+\frac{x}{z}& =& -\frac{1}{2}+\frac{\sqrt{3}}{2}\cot \theta -\frac{1}{2}-\frac{\sqrt{3}}{2}\cot \theta \\ \Rightarrow \frac{zx+xy}{yz}& =& -1\\ \Rightarrow zx+xy& =& -yz\\ & \Rightarrow & zx+xy+yz=0\end{array}$

Hence, option C is correct.