If $\cos α + \cos β + \cos γ = \sin α + \sin β + \sin γ = 0 then \cos (β + γ) + \cos (α + β) + \cos (α + γ)$ is equal to

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$\frac{1}{2}$

answer is B.

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Detailed Solution

Let $a = \cos α + i \sin α; b = \cos β + i + \sin β and c =$ $\cos γ + i \sin γ$ then $a + b + c = 0$ and $| a | = | b | = | c | = 1$

$\begin{array}{l} ∴ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \bar{a} + \bar{b} + \bar{c} \\ \bar{a} + \bar{b} + \bar{c} = (\cos α + \cos β + \cos γ) - i (\sin α + \\ \sin β + \sin γ) = 0 \\ (\begin{matrix} ∴ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \\ ∴ a b + b c + c a = 0 \end{matrix}) ab + bc + ac = 0 \\ \Rightarrow [\cos (α + β) + \cos (β + γ) + \cos (α + γ)] \\ + i [\sin (α + β) + \sin (α + γ) + \sin (β + γ)] = 0 \\ \Rightarrow \cos (α + β) + \cos (β + γ) + \cos (α + γ) = 0 \end{array}$