If $x \cos heta=y \cos \left( heta+\frac{2 \pi}{3} ight)=z \cos \left( heta+\frac{4 \pi}{3} ight)$, then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=$

Question

If $x \cos 	heta=y \cos \left(	heta+\frac{2 \pi}{3}ight)=z \cos \left(	heta+\frac{4 \pi}{3}ight)$, then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=$

Aicharya · Accepted Answer

$ \begin{aligned} & 	ext { Given, } x \cos 	heta=y \cos \left(	heta+\frac{2 \pi}{3}ight) \ & z \cos \left(	heta+\frac{4 \pi}{3}ight)=\lambda 	ext { (say) } \ & \Rightarrow \cos 	heta=\frac{\lambda}{x}, \cos \left(	heta+\frac{2 \pi}{3}ight)=\frac{\lambda}{y}, \cos \left(	heta+\frac{4 \pi}{3}ight)=\frac{\lambda}{z} \ & 	herefore \cos 	heta+\cos \left(	heta+\frac{2 \pi}{3}ight)+\cos \left(	heta+\frac{4 \pi}{3}ight) \ & =\lambda\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}ight) \ & \Rightarrow \lambda\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}ight)=2 \cos \left(\frac{	heta+	heta+4 \pi / 3}{2}ight) \ & \cos \left(\frac{	heta+4 \pi / 3-	heta}{2}ight)+\cos \left(	heta+\frac{2 \pi}{3}ight) \ & \Rightarrow \quad\left[\because \cos A+\cos B=2 \cos \left(\frac{A+B}{2}ight) \cos \left(\frac{A-B}{2}ight)ight] \ & \Rightarrow \quad \lambda\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}ight)=2 \cos \left(	heta+\frac{2 \pi}{3}ight) \ & \Rightarrow \quad \cos \left(\frac{2 \pi}{3}ight)+\cos \left(	heta+\frac{2 \pi}{3}ight) \end{aligned} $ $ \begin{aligned} & \because \cos \left(\frac{2 \pi}{3}ight)=\cos \left(\pi-\frac{\pi}{3}ight)=-\cos \frac{\pi}{3}=-\frac{1}{2} \ & 	herefore \lambda\left(\frac{1}{x}+\frac{1}{y}+\frac{+1}{z}ight)=-\cos \left(	heta+\frac{2 \pi}{3}ight)+\cos \left(	heta+\frac{2 \pi}{3}ight) \ & =0 \ & 	herefore \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0 \quad \quad[\because \lambda 
eq 0] \end{aligned} $

If $x \cos \theta=y \cos \left(\theta+\frac{2 \pi}{3}\right)=z \cos \left(\theta+\frac{4 \pi}{3}\right)$,…

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