In triangle ABC, frac{1 + cos(A - B)cos C}{1 | vedclass

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(D) We know that in a triangle $ABC,$ $C = 180^\circ - (A + B),$ so $\cos C = -\cos(A + B).$Substituting this into the numerator: $1 + \cos(A - B)\cos

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$\frac{a^2 + b^2}{a^2 + c^2}$

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(D) We know that in a triangle $ABC,$ $C = 180^\circ - (A + B),$ so $\cos C = -\cos(A + B).$Substituting this into the numerator: $1 + \cos(A - B)\cos C = 1 - \cos(A - B)\cos(A + B).$Using the identity $\cos(A - B)\cos(A + B) = \cos^2 A - \sin^2 B,$ we get $1 - (\cos^2 A - \sin^2 B) = 1 - \cos^2 A + \sin^2 B = \sin^2 A + \sin^2 B.$Similarly,for the denominator: $1 + \cos(A - C)\cos B = 1 - \cos(A - C)\cos(A + C) = 1 - (\cos^2 A - \sin^2 C) = \sin^2 A + \sin^2 C.$Thus,the expression becomes $\frac{\sin^2 A + \sin^2 B}{\sin^2 A + \sin^2 C}.$Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k,$ we have $\sin A = \frac{a}{k}, \sin B = \frac{b}{k}, \sin C = \frac{c}{k}.$Substituting these values,we get $\frac{(a/k)^2 + (b/k)^2}{(a/k)^2 + (c/k)^2} = \frac{a^2 + b^2}{a^2 + c^2}.$

In triangle $ABC,$ $\frac{1 + \cos(A - B)\cos C}{1 + \cos(A - C)\cos B} = $

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