In a triangle ABC, if $Cos A + 2 Cos B + Cos C = 2$ and the lengths of the sides opposite to the angles A and C are 3 and 7 respectively, then $\cos A - \cos C$ is equal to

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$\frac{10}{7}$

$\frac{9}{7}$

$\frac{3}{7}$

$\frac{5}{7}$

answer is A.

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Detailed Solution

$\begin{array}{l} \cos A + 2 \cos B + \cos C = 2 \\ \cos A + \cos C = 2 (1 - \cos B) \\ 2 \cos \frac{A + C}{2} \cos (\frac{A - C}{2}) = 2 \times 2 \sin^{2} \frac{B}{2} \\ \cos \frac{A - C}{2} = 2 \sin \frac{B}{2} \\ 2 \cos \frac{B}{2} \cos \frac{A - C}{2} = 4 \sin \frac{B}{2} \cos \frac{B}{2} \\ 2 \sin (\frac{A + C}{2}) \cos (\frac{A - C}{2}) = 2 \sin B \\ \sin A + \sin C = 2 \sin B \frac{a}{2 R} + \frac{c}{2 R} = \frac{2 b}{2 R} \end{array}$
$\begin{array}{l} a + c = 2 b \\ \Rightarrow b = 5 (∵ a = 3, c = 7) \\ \cos A - \cos C = = \frac{b^{2} + c^{2} - a^{2}}{2 b c} - \frac{a^{2} + b^{2} - c^{2}}{2 a b} \frac{25 + 49 - 9}{70} - \frac{9 + 25 - 49}{30} = \frac{10}{7} \end{array}$