In any triangle ABC,the expression frac{1+cos | vedclass

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(D) Given the expression $\frac{1+\cos(A-B) \cdot \cos C}{1+\cos(A-C) \cdot \cos B}$.Since $A+B+C = 180^{\circ}$,we have $C = 180^{\circ} - (A+B)$ and

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

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(D) Given the expression $\frac{1+\cos(A-B) \cdot \cos C}{1+\cos(A-C) \cdot \cos B}$.Since $A+B+C = 180^{\circ}$,we have $C = 180^{\circ} - (A+B)$ and $B = 180^{\circ} - (A+C)$.Thus,$\cos C = -\cos(A+B)$ and $\cos B = -\cos(A+C)$.Substituting these into the expression:$= \frac{1 - \cos(A-B)\cos(A+B)}{1 - \cos(A-C)\cos(A+C)}$Using the identity $\cos(x-y)\cos(x+y) = \cos^2 x - \sin^2 y$:$= \frac{1 - (\cos^2 A - \sin^2 B)}{1 - (\cos^2 A - \sin^2 C)}$$= \frac{1 - \cos^2 A + \sin^2 B}{1 - \cos^2 A + \sin^2 C}$$= \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 $\sin A = \frac{a}{k}, \sin B = \frac{b}{k}, \sin C = \frac{c}{k}$.$= \frac{(a/k)^2 + (b/k)^2}{(a/k)^2 + (c/k)^2} = \frac{a^2+b^2}{a^2+c^2}$.

In any $\triangle ABC$,the expression $\frac{1+\cos(A-B) \cdot \cos C}{1+\cos(A-C) \cdot \cos B}$ is equal to:

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