Question

If x cos theta=ycos(theta+(2pi/3))=zcos{theta+(4pi/3)}. find xy+yz+xz

Answer 1

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

Therefore, $\frac{1}{k} = x \cos \theta$ , $\frac{1}{k} = y \cos(\theta + \frac{2 \pi}{3})$ and $\frac{1}{k} = z \cos(\theta + \frac{4 \pi}{3})$

So, $\frac{1}{x} = k \cos \theta$ , $k \cos(\theta + \frac{2 \pi}{3}) = \frac{1}{y}$ and $\frac{1}{z} = k \cos(\theta + \frac{4 \pi}{3})$.

Consider $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$

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

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

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

By using the trigonometric identity

$\cos A + \cos B = 2 \cos(\frac{A-B}{2}) \cos(\frac{A+B}{2})$

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

= $k(2 \cos(\frac{\pi- \pi}{3}) \cos(\frac{\theta+2 \pi}{3}) + \cos(\frac{\theta+2 \pi}{3}))$

= $k(-2 \cos\frac{\pi}{3} \cos(\frac{\theta+2 \pi}{3}) + \cos(\frac{\theta+2 \pi}{3}))$

= $k(-2 \times \frac{1}{2} \cos(\frac{\theta+2 \pi}{3}) + \cos(\frac{\theta+2 \pi}{3}))$

= $k(-\cos(\frac{\theta+2 \pi}{3}) + \cos(\frac{\theta+2 \pi}{3}))$

= 0

Therefore,

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

Taking LCM as xyz

$\frac{yz+xz+xy}{xyz} = 0$

So, xy+yz+xz = 0

Hence, proved.

Answer 2

hope this will help u