Question

If If $xcos\theta =ycos(\theta +\frac{2\pi }{3})=zcos(\theta +\frac{4\pi }{3}),$ then write the value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$.

Answer 1

Let, $xcos\theta =ycos(\theta +\frac{2\pi }{3})=zcos(\theta +\frac{4\pi }{3})=k$
$\Rightarrow xcos\theta =k,ycos(\theta +\frac{2\pi }{3})=k$ and,
$zcos(\theta +\frac{4\pi }{3})=k$
$\Rightarrow \frac{1}{x}=\frac{cos\theta }{k},\frac{1}{y}=\frac{cos(\frac{2\pi }{3}+\theta )}{k}$ <br/>
and, $\frac{1}{z}=\frac{cos(\frac{4\pi }{3}+\theta )}{k}$<br/>
Now, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{cos\theta }{k}+\frac{cos(\frac{2\pi }{3}+\theta )}{k}+\frac{cos(\frac{4\pi }{3}+\theta )}{k}$<br/>
$\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{k}[cos\theta +cos(\frac{2\pi }{3}+\theta )+cos(\frac{4\pi }{3}+\theta )]$<br/>
$=\frac{1}{k}[cos\theta +cos\{\frac{\pi }{2}+(\frac{\pi }{6}+\theta )\}cos\{\pi +(\frac{\pi }{3}+\theta )\}]=\frac{1}{k}[cos\theta -sin(\frac{\pi }{6}+\theta )-cos(\frac{\pi }{3}+\theta )]$<br/>
$=\frac{1}{k}[cos\theta -(sin\frac{\pi }{6}cos\theta +cos\frac{\pi }{6}sin\theta )-(cos\frac{\pi }{3}\times cos\theta -sin\frac{\pi }{3}\times sin\theta )]$<br/>
$=\frac{1}{k}[cos\theta -(\frac{1}{2}cos\theta +\frac{\sqrt{3}}{2}sin\theta )-(\frac{1}{2}cos\theta -\frac{\sqrt{3}}{2}sin\theta )]$<br/>
$=\frac{1}{k}[cos\theta -\frac{1}{2}cos\theta -\frac{\sqrt{3}}{2}sin\theta -\frac{1}{2}cos\theta +\frac{\sqrt{3}}{2}sin\theta ]$<br/>
=0