If $A + B + C = 2 k$ , then prove that ${cos}^{2} k + {cos}^{2} (k - A) + {cos}^{2} (k - B) + {cos}^{2} (k - C) = 2 + 2 cos A cos B cos C$ .

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Solution

${cos}^{2} x - {sin}^{2} y = cos (x + y) cos (x - y)$
L.H.S. $= 2 + [{cos}^{2} k - {sin}^{2} (k - A)] + [{cos}^{2} (k - B) - {sin}^{2} (k - C)]$
$= 2 + cos (2 k - A) cos A + cos (2 k - B - C) cos (C - B)$
Put $2 k - A = B + C$ and $2 k - B - C = A$
$= 2 + cos A [cos (B + C) + cos (C - B)]$
$= 2 + 2 cos A cos B cos C$ .

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