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$\cos (α + β + γ) + \cos (α - β - γ) + \cos (β - γ - α) + \cos (γ - α - β) =$

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4 $\cos αcos βcos γ$

6 $\cos αcos βcos γ$

3 $\cos αcos βcos γ$

2 $\cos αcos βcos γ$

answer is C.

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Detailed Solution

$I = \cos (α + β + γ) + \cos (α - β - γ) + \cos (β - γ - α) + \cos (γ - α - β)$

$= \cos (γ + (α + β)) + \cos (γ - (α + β)) + \cos ((α - β) - γ) + \cos ((α - β) + γ)$

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

we get $I = 2 \cos γ \cos (α + β) + 2 \cos (α - β) \cos γ$

$= 2 \cos γ (\cos (α + β) + \cos (α - β))$

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

$I = 2 \cos γ (\cos (α + β) + \cos (α - β))$

$\Rightarrow I = 2 \cos γ (2 \cos α \cos β) = 4 \cos α \cos β \cos γ$

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