In a triangle ABC,if cos A + 2 cos B + cos C | vedclass

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(C) Given $\cos A + \cos C = 2(1 - \cos B)$.Using the sum-to-product formula,$2 \cos \frac{A+C}{2} \cos \frac{A-C}{2} = 4 \sin^2 \frac{B}{2}$.Since $\

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$\frac{10}{7}$

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(C) Given $\cos A + \cos C = 2(1 - \cos B)$.Using the sum-to-product formula,$2 \cos \frac{A+C}{2} \cos \frac{A-C}{2} = 4 \sin^2 \frac{B}{2}$.Since $\cos \frac{A+C}{2} = \sin \frac{B}{2}$,we have $2 \sin \frac{B}{2} \cos \frac{A-C}{2} = 4 \sin^2 \frac{B}{2}$,which simplifies to $\cos \frac{A-C}{2} = 2 \sin \frac{B}{2}$.Using the sine rule,$\sin A + \sin C = 2 \sin B$,which implies $a + c = 2b$.Given $a = 3$ and $c = 7$,we get $3 + 7 = 2b$,so $b = 5$.Now,$\cos A - \cos C = \frac{b^2 + c^2 - a^2}{2bc} - \frac{a^2 + b^2 - c^2}{2ab}$.Substituting the values: $\cos A - \cos C = \frac{25 + 49 - 9}{2(5)(7)} - \frac{9 + 25 - 49}{2(3)(5)}$.$= \frac{65}{70} - \frac{-15}{30} = \frac{13}{14} + \frac{1}{2} = \frac{13 + 7}{14} = \frac{20}{14} = \frac{10}{7}$.

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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