Cosc/cosb=sinb/sinc
sin2b=sin2c
either 2b=2c
or 2b+2c=π so A = π/2

Given:
In triangle ABC,
$\frac{\cos A \cdot 2\cos C}{\cos A \cdot 2\cos B} = \frac{\sin B}{\sin C}$

To prove:
Triangle ABC is either isosceles or right-angled.

Proof:

1. Simplifying the given equation:
Using the identity $\sin B = 2\sin\frac{B}{2}\cos\frac{B}{2}$ and $\cos C = 2\cos^2\frac{C}{2} - 1$, we can simplify the given equation as follows:

$\frac{\cos A \cdot 2\cos C}{\cos A \cdot 2\cos B} = \frac{\sin B}{\sin C}$

$\frac{\cos A \cdot 2(2\cos^2\frac{C}{2} - 1)}{\cos A \cdot 2\cos B} = \frac{2\sin\frac{B}{2}\cos\frac{B}{2}}{2\sin\frac{C}{2}\cos\frac{C}{2}}$

Cancelling out common factors:

$\frac{2\cos^2\frac{C}{2} - 1}{\cos B} = \frac{\sin\frac{B}{2}}{\sin\frac{C}{2}}$

2. Applying the Sine Rule:
Using the Sine Rule in triangle ABC, we have:

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

Therefore, we can rewrite the equation as:

$\frac{2\cos^2\frac{C}{2} - 1}{\cos B} = \frac{\sin\frac{B}{2}}{\sin\frac{C}{2}} = \frac{a}{c}$

3. Applying the Cosine Rule:
Using the Cosine Rule in triangle ABC, we have:

$c^2 = a^2 + b^2 - 2ab\cos C$

Simplifying the equation, we get:

$2ab\cos C = a^2 + b^2 - c^2$

Substituting the value of $\cos C$ from the given equation, we have:

$2ab(2\cos^2\frac{C}{2} - 1) = a^2 + b^2 - c^2$

Simplifying further, we get:

$4ab\cos^2\frac{C}{2} - 2ab = a^2 + b^2 - c^2$

4. Analyzing the equation:
From the given equation, we have:

$\frac{2\cos^2\frac{C}{2} - 1}{\cos B} = \frac{a}{c}$

Rearranging the equation, we get:

$2\cos^2\frac{C}{2} - 1 = \frac{a}{c}\cos B$

Replacing $\cos^2\frac{C}{2}$ with $\frac{1 + \cos C}{2}$ (from the half