Let's analyze the given options using the properties of triangles. Option A: \( \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \) This is the standard Sine Rule, which is universally true for any triangle. Option B: \( \frac{a-b}{a+b}=\frac{\sin A-\sin B}{\sin A+\sin B} \) Using the Sine Rule, we can substitute \( a = k \sin A \) and \( b = k \sin B \). \[ \frac{a-b}{a+b} = \frac{k \sin A - k \sin B}{k \sin A + k \sin B} = \frac{\sin A - \sin B}{\sin A + \sin B} \] Thus, this statement is correct. Option C: \( \frac{a-b}{a+b}=\cot\left(\frac{A+B}{2}\right)\tan\left(\frac{A-B}{2}\right) \) Starting from the result in Option B: \[ \frac{\sin A - \sin B}{\sin A + \sin B} = \frac{2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)}{2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)} \] \[ = \cot\left(\frac{A+B}{2}\right)\tan\left(\frac{A-B}{2}\right) \] This is Napier's Analogy (Law of Tangents), meaning this statement is also correct. Option D: \( \frac{a-b}{a+b}=\frac{\tan\left(\frac{A-B}{2}\right)}{\tan\left(\frac{C}{2}\right)} \) We know that \( A + B + C = \pi \), so \( \frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2} \). Therefore, \( \cot\left(\frac{A+B}{2}\right) = \cot\left(\frac{\pi}{2} - \frac{C}{2}\right) = \tan\left(\frac{C}{2}\right) \). Substitute this into the expression from Option C: \[ \frac{a-b}{a+b} = \tan\left(\frac{C}{2}\right)\tan\left(\frac{A-B}{2}\right) \] The statement in Option D expresses this as division by \( \tan\left(\frac{C}{2}\right) \) instead of multiplication. Hence, Option D is the incorrect mathematical statement.