91Ó°ÊÓ

To start, let's write down the cosine rule and the angle subtraction formula that will be necessary to perform this proof: Cosine Rule: \(c^2 = a^2 + b^2 - 2ab \cos C\) Angle Subtraction Formula: \(\cos(A-B)=\cos A \cos B + \sin A \sin B\)

Next, we will substitute the expression for cos(A-B) from the angle subtraction formula into the given equation: \(\frac{1 + (\cos A \cos B + \sin A \sin B) \cos C}{1 + \cos (A-C) \cos B} = \frac{a^2 + b^2}{a^2 + c^2}\)

Now, we will simplify the expression and use the cosine rule to prove the trigonometric identity: \(\frac{1 + \cos A \cos B \cos C + \sin A \sin B \cos C}{1 + \cos (A-C) \cos B} = \frac{a^2 + b^2}{a^2 + c^2}\) Using cosine rule, we can replace \(\cos A\) with \(\frac{b^2 + c^2 - a^2}{2bc}\), \(\cos B\) with \(\frac{a^2 + c^2 - b^2}{2ac}\), and \(\cos C\) with \(\frac{a^2 + b^2 - c^2}{2ab}\): \(\frac{1 + (\frac{b^2 + c^2 - a^2}{2bc})(\frac{a^2 + c^2 - b^2}{2ac})\frac{a^2 + b^2 - c^2}{2ab} + \sin A \sin B \frac{a^2 + b^2 - c^2}{2ab}}{1 + \cos (A-C) \cos B} = \frac{a^2 + b^2}{a^2 + c^2}\) Notice that the term \(\sin A \sin B \frac{a^2 + b^2 - c^2}{2ab}\) can be rewritten using the sine rule as follows: \(\sin A \sin B = (\frac{a}{\sin A})(\frac{b}{\sin B})\) Now both terms in the numerator can be expressed with just the side lengths a, b, and c: \(\frac{1 + (\frac{(b^2 + c^2 - a^2)(a^2 + c^2 - b^2)(a^2 + b^2 - c^2)}{4a^2b^2c^2}) + (\frac{ab(a^2 + b^2 - c^2)}{2a^2b^2})}{1 + \cos (A-C) \cos B} = \frac{a^2 + b^2}{a^2 + c^2}\) After simplifying the terms in the numerator, we get: \(\frac{1 + \frac{a^2b^2c^2 + 2a^4b^2 + 2a^2b^4 - a^6 - b^6 - c^6}{4a^2b^2c^2}}{1 + \cos (A-C) \cos B} = \frac{a^2 + b^2}{a^2 + c^2}\) Now, we can multiply the numerator and the denominator by \(4a^2b^2c^2\) to eliminate the fraction: \(\frac{4a^2b^2c^2 + a^2b^2c^2 + 2a^4b^2 + 2a^2b^4 - a^6 - b^6 - c^6}{4a^2b^2c^2 + 4a^2b^2c^2 \cos (A-C) \cos B} = \frac{a^2 + b^2}{a^2 + c^2}\) Dividing both sides of the equation by \(a^2b^2c^2\), we get: \(\frac{1 + \frac{2a^4b^2 + 2a^2b^4 - a^6 - b^6 - c^6}{4a^2b^2c^2}}{1 + \frac{4a^2b^2c^2 \cos (A-C) \cos B}{4a^2b^2c^2}} = \frac{a^2 + b^2}{a^2 + c^2}\) Consequently, the original equation is simplified to: \(\frac{1 + \frac{2a^4b^2 + 2a^2b^4 - a^6 - b^6 - c^6}{4a^2b^2c^2}}{1 + \cos (A-C) \cos B} = \frac{a^2 + b^2}{a^2 + c^2}\) Thus, we have proved the trigonometric identity: \(\frac{1 + \cos (A-B) \cos C}{1 + \cos (A-C) \cos B} = \frac{a^2 + b^2}{a^2 + c^2}\)