Question

If x cos θ = y cos $\left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)=z\cos \left(\mathrm{\theta }+\frac{4\mathrm{\pi }}{3}\right)$, then write the value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$.

Answer 1

$$\begin{gathered} \text{Given}: \\ x\cos \theta =y\left(\cos \theta \cos \frac{2\mathrm{\pi }}{3}-\sin \theta \sin \frac{2\mathrm{\pi }}{3}\right)=z\left(\cos \theta \cos \frac{4\mathrm{\pi }}{3}-\sin \theta \sin \frac{4\mathrm{\pi }}{3}\right) \\ \Rightarrow x\cos \theta =y\left(-\frac{1}{2}\cos \theta -\frac{\sqrt{3}}{2}\sin \theta \right)=z\left(-\frac{1}{2}\cos \theta +\frac{\sqrt{3}}{2}\sin \theta \right) \\ \Rightarrow x=\frac{y}{2}\left(-1-\sqrt{3}\tan \theta \right)=\frac{z}{2}\left(-1+\sqrt{3}\tan \theta \right) \\ x=\frac{y}{2}\left(-1-\sqrt{3}\tan \theta \right) \\ z=\frac{y\left(-1-\sqrt{3}\tan \theta \right)}{\left(-1+\sqrt{3}\tan \theta \right)} \\ \mathrm{Now}, \\ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{2}{y\left(-1-\sqrt{3}\tan \theta \right)}+\frac{1}{y}+\frac{\left(-1+\sqrt{3}\tan \theta \right)}{y\left(-1-\sqrt{3}\tan \theta \right)} \\ =\frac{2+\left(-1-\sqrt{3}\tan \theta \right)+\left(-1+\sqrt{3}\tan \theta \right)}{y\left(-1-\sqrt{3}\tan \theta \right)} \\ =0 \end{gathered}$$