If xcos⁡θ=ycos⁡θ+2π3=zcos⁡θ+4π3,then the value of 1x+1y+1z is equal to

A
1
B
2
C
0
D
3 cos $θ$

Solution:

We have, $xcos θ = ycos (θ + \frac{2 π}{3})$ $= zcos (θ + \frac{4 π}{3}) = k$

$\begin{aligned} \begin{array}{ll} \Rightarrow \cos θ = \frac{k}{x}, \cos (θ + \frac{2 π}{3}) = \frac{k}{y} and \cos (θ + \frac{4 π}{3}) = \frac{k}{z} \\ \begin{aligned} \frac{k}{x} + \frac{k}{y} + \frac{k}{z} & = \cos θ + \cos (θ + \frac{2 π}{3}) + \cos (θ + \frac{4 π}{3}) \\ = \cos θ - \cos (\frac{π}{3} - θ) - \cos (\frac{π}{3} + θ) \end{aligned} \\ = \cos θ - 2 \cos \frac{π}{3} \cos θ = 0 \\ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \end{array} \end{aligned}$

If $xcos θ = ycos (θ + \frac{2 π}{3}) = zcos (θ + \frac{4 π}{3})$ ,then the value of $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ is equal to

Solution:

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