Questions

If $xsin θ = ysin (θ + \frac{2 π}{3}) = zsin (θ + \frac{4 π}{3})$ then

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x+y +z = 0

xy + yz + zx = 0

xyz +x+y +z = 1

none of these

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detailed solution

Correct option is B

we have,

$\begin{aligned} xsin θ = ysin (θ + \frac{2 π}{3}) = zsin (θ + \frac{4 π}{3}) \\ \Rightarrow \frac{\sin θ}{\frac{1}{x}} = \frac{\sin (θ + \frac{2 π}{3})}{\frac{1}{y}} = \frac{\sin (θ + \frac{4 π}{3})}{\frac{1}{z}} \\ \Rightarrow \frac{\sin θ}{\frac{1}{x}} = \frac{\sin (θ + \frac{2 π}{3})}{\frac{1}{y}} = \frac{\sin (θ + \frac{4 π}{3})}{\frac{1}{z}} \\ = \frac{\sin θ + \sin (θ + \frac{2 π}{3}) + \sin (θ + \frac{4 π}{3})}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \\ \Rightarrow \frac{\sin θ}{\frac{1}{x}} = \frac{\sin (θ + \frac{2 π}{3})}{\frac{1}{y}} = \frac{\sin (θ + \frac{4 π}{3})}{\frac{1}{z}} \\ = \frac{\sin θ + 2 \sin (π + θ) \cos \frac{π}{3}}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \\ \Rightarrow \frac{\sin θ}{\frac{1}{x}} \times (\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) = 0 \\ \Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \Rightarrow xy + yz + zx = 0 \end{aligned}$