Question

If x \cos \theta=y \cos \left(\theta+\frac{2 \pi}{3}\right)=z \cos \left(\theta+\frac{4 \pi}{3}\right)xcosθ=ycos(θ+ 3 2π ​ )=zcos(θ+ 3 4π ​ ) then x y+y z+z xxy+yz+zx equals to

Answer 1

What is this question

Answer 2

SOLUTION

GIVEN

$\displaystyle \sf{}x \cos \theta = y \cos \bigg(\theta + \frac{2\pi}{3} \bigg) =z \cos \bigg(\theta + \frac{4\pi}{3} \bigg)$

TO DETERMINE

$\sf{}xy + yz + zx$

EVALUATION

Let

$\displaystyle \sf{}x \cos \theta = y \cos \bigg(\theta + \frac{2\pi}{3} \bigg) =z \cos \bigg(\theta + \frac{4\pi}{3} \bigg) = \frac{1}{k} (say)$

$\displaystyle \sf{} \frac{1}{x} = k \cos \theta$

$\displaystyle \sf{} \frac{1}{y} = k \cos \bigg(\theta + \frac{2\pi}{3} \bigg)$

$\displaystyle \sf{} \frac{1}{z} = k\cos \bigg(\theta + \frac{4\pi}{3} \bigg)$

So

$\displaystyle \sf{} \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$

$\displaystyle \sf{} = k \bigg(\cos \theta + \cos \big(\theta + \frac{2\pi}{3} \big) + \cos \big(\theta + \frac{4\pi}{3} \big) \bigg)$

$\displaystyle \sf{} = k \cos \theta +2k \cos (\theta + \pi ) .\cos \frac{\pi}{3}$

$\displaystyle \sf{} = k \cos \theta - 2k \cos \theta . \frac{1}{2}$

$\displaystyle \sf{} = k \cos \theta - k \cos \theta$

$= \sf{}0$

So

$\displaystyle \sf{} \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$

$\implies \displaystyle \sf{} \frac{yz + zx + xy}{xyz} = 0$

$\implies \displaystyle \sf{} {yz + zx + xy} = 0$

$\implies \sf{}xy + yz + zx = 0$

FINAL ANSWER

$\boxed{ \sf{} \: \: xy + yz + zx = 0 \: \: }$

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