Step 1: Use the angle sum property of a triangle. In any triangle ABC, the sum of angles is $A+B+C = 180^\circ$. Therefore, $B = 180^\circ - (A+C)$. Using the cosine identity $\cos(180^\circ - \theta) = -\cos \theta$, we have: $$ \cos B = \cos(180^\circ - (A+C)) = -\cos(A+C) $$ Step 2: Substitute this into the given equation. The given equation is $\frac{\sin A + \cos A}{\cos B} = \sqrt{2}$. Substitute $\cos B = -\cos(A+C)$: $$ \frac{\sin A + \cos A}{-\cos(A+C)} = \sqrt{2} $$ $$ \sin A + \cos A = -\sqrt{2} \cos(A+C) $$ Step 3: Expand $\cos(A+C)$ and rearrange the terms. Using the cosine addition formula $\cos(X+Y) = \cos X \cos Y - \sin X \sin Y$: $$ \sin A + \cos A = -\sqrt{2} (\cos A \cos C - \sin A \sin C) $$ $$ \sin A + \cos A = -\sqrt{2} \cos A \cos C + \sqrt{2} \sin A \sin C $$ Rearrange the terms to group $\sin A$ and $\cos A$: $$ \sin A - \sqrt{2} \sin A \sin C + \cos A + \sqrt{2} \cos A \cos C = 0 $$ Factor out $\sin A$ and $\cos A$: $$ \sin A (1 - \sqrt{2} \sin C) + \cos A (1 + \sqrt{2} \cos C) = 0 $$ Step 4: Determine the values of $\sin C$ and $\cos C$. For the equation $\sin A (1 - \sqrt{2} \sin C) + \cos A (1 + \sqrt{2} \cos C) = 0$ to hold true for any angle $A$ in a triangle, the coefficients of $\sin A$ and $\cos A$ must both be zero, as $\sin A$ and $\cos A$ are linearly independent. Thus, we must have: $$ 1 - \sqrt{2} \sin C = 0 \quad \text{(Equation 1)} $$ $$ 1 + \sqrt{2} \cos C = 0 \quad \text{(Equation 2)} $$ From Equation 1: $$ \sqrt{2} \sin C = 1 \implies \sin C = \frac{1}{\sqrt{2}} $$ From Equation 2: $$ \sqrt{2} \cos C = -1 \implies \cos C = -\frac{1}{\sqrt{2}} $$ Step 5: Find the angle $C$. We need an angle $C$ such that $\sin C = \frac{1}{\sqrt{2}}$ and $\cos C = -\frac{1}{\sqrt{2}}$. Since $\sin C$ is positive and $\cos C$ is negative, angle $C$ must be in the second quadrant. The reference angle for which $\sin \theta = \frac{1}{\sqrt{2}}$ and $\cos \theta = \frac{1}{\sqrt{2}}$ is $45^\circ$. In the second quadrant, the angle is $180^\circ - 45^\circ = 135^\circ$. Therefore, $C = 135^\circ$. This value is valid for an angle in a triangle ($0^\circ < C < 180^\circ$). Thus, we have proven that $C = 135^\circ$. The final answer is $\boxed{\text{C = 135^\circ}}$