Question

If $x\cos \theta =y\cos \left(\theta +\frac{2\mathrm{\pi }}{3}\right)=z\cos \left(\theta +\frac{4\mathrm{\pi }}{3}\right)$, prove that $xy+yz+zx=0$. [NCERT EXEMPLAR]

Answer 1

$$\begin{gathered} x\cos \theta =y\cos \left(\theta +\frac{2\mathrm{\pi }}{3}\right)=z\cos \left(\theta +\frac{4\mathrm{\pi }}{3}\right) \\ \Rightarrow \frac{\cos \theta }{{\displaystyle \frac{1}{x}}}=\frac{\cos \left(\theta +{\displaystyle \frac{2\mathrm{\pi }}{3}}\right)}{{\displaystyle \frac{1}{y}}}=\frac{\cos \left(\theta +{\displaystyle \frac{4\mathrm{\pi }}{3}}\right)}{{\displaystyle \frac{1}{z}}} \\ \Rightarrow \frac{\cos \theta }{{\displaystyle \frac{1}{x}}}=\frac{\cos \left(\theta +{\displaystyle \frac{2\mathrm{\pi }}{3}}\right)}{{\displaystyle \frac{1}{y}}}=\frac{\cos \left(\theta +{\displaystyle \frac{4\mathrm{\pi }}{3}}\right)}{{\displaystyle \frac{1}{z}}}=\frac{\cos \theta +\cos \left(\theta +{\displaystyle \frac{2\mathrm{\pi }}{3}}\right)+\cos \left(\theta +{\displaystyle \frac{4\mathrm{\pi }}{3}}\right)}{{\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{y}}+{\displaystyle \frac{1}{z}}}\left(\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=...=\frac{a+c+e+...}{b+d+f+...}\right) \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{\cos \theta }{{\displaystyle \frac{1}{x}}}=\frac{\cos \left(\theta +{\displaystyle \frac{2\mathrm{\pi }}{3}}\right)}{{\displaystyle \frac{1}{y}}}=\frac{\cos \left(\theta +{\displaystyle \frac{4\mathrm{\pi }}{3}}\right)}{{\displaystyle \frac{1}{z}}}=\frac{\cos \theta +2\cos \left({\displaystyle \frac{\theta +{\displaystyle \frac{2\mathrm{\pi }}{3}+}\theta +{\displaystyle \frac{4\mathrm{\pi }}{3}}}{2}}\right)\cos \left({\displaystyle \frac{\theta +{\displaystyle \frac{2\mathrm{\pi }}{3}-}\theta -{\displaystyle \frac{4\mathrm{\pi }}{3}}}{2}}\right)}{{\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{y}}+{\displaystyle \frac{1}{z}}} \\ \Rightarrow \frac{\cos \theta }{{\displaystyle \frac{1}{x}}}=\frac{\cos \left(\theta +{\displaystyle \frac{2\mathrm{\pi }}{3}}\right)}{{\displaystyle \frac{1}{y}}}=\frac{\cos \left(\theta +{\displaystyle \frac{4\mathrm{\pi }}{3}}\right)}{{\displaystyle \frac{1}{z}}}=\frac{\cos \theta +2\cos \left({\displaystyle \mathrm{\pi }+\theta }\right)\cos \left({\displaystyle -\frac{\mathrm{\pi }}{3}}\right)}{{\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}}} \\ \Rightarrow \frac{\cos \theta }{{\displaystyle \frac{1}{x}}}=\frac{\cos \left(\theta +{\displaystyle \frac{2\mathrm{\pi }}{3}}\right)}{{\displaystyle \frac{1}{y}}}=\frac{\cos \left(\theta +{\displaystyle \frac{4\mathrm{\pi }}{3}}\right)}{{\displaystyle \frac{1}{z}}}=\frac{\cos \theta +2\times \left(-\cos \theta \right)\times {\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}}}\left[\cos \left(-\theta \right)=\cos \theta \right] \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{\cos \theta }{{\displaystyle \frac{1}{x}}}=\frac{\cos \left(\theta +{\displaystyle \frac{2\mathrm{\pi }}{3}}\right)}{{\displaystyle \frac{1}{y}}}=\frac{\cos \left(\theta +{\displaystyle \frac{4\mathrm{\pi }}{3}}\right)}{{\displaystyle \frac{1}{z}}}=\frac{0}{{\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}}} \\ \Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0 \\ \Rightarrow \frac{yz+zx+xy}{xyz}=0 \\ \Rightarrow xy+yz+zx=0 \end{gathered}$$