If xcos theta = ycos ,left( {theta + frac{{2p | vedclass

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(c) We have
$x\cos 	heta = y\cos \left( {	heta + \frac{{2\pi }}{3}} ight) = z\cos \left( {	heta + \frac{{4\pi }}{3}} ight) = k$
==> $\cos

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(c) We have
$x\cos 	heta = y\cos \left( {	heta + \frac{{2\pi }}{3}} ight) = z\cos \left( {	heta + \frac{{4\pi }}{3}} ight) = k$
==> $\cos 	heta = \frac{k}{x},\cos \left( {	heta + \frac{{2\pi }}{3}} ight) = \frac{k}{y}$
and $\cos \left( {	heta + \frac{{4\pi }}{3}} ight) = \frac{k}{z}$
Hence $\frac{k}{x} + \frac{k}{y} + \frac{k}{z} = \cos 	heta + \cos \left( {	heta + \frac{{2\pi }}{3}} ight) + \cos \left( {	heta + \frac{{4\pi }}{3}} ight)$
$ = \cos 	heta + \cos \left( {\frac{\pi }{3} - 	heta } ight) - \cos \left( {\frac{\pi }{3} + 	heta } ight)$
$ = \cos 	heta - 2\cos \frac{\pi }{3}\cos 	heta = 0$.

If $x\cos \theta = y\cos \,\left( {\theta + \frac{{2\pi }}{3}} \right) = z\cos \,\left( {\theta + \frac{{4\pi }}{3}} \right),$ then the value of $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ is equal to

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