(b) \( \frac{\sin A}{\sin B}-\frac{\cos A}{\cos B}=\frac{2 \sin (A-B)}{\sin 2 B} \)

1. Start with the left-hand side (LHS): \[ \frac{\sin A}{\sin B} - \frac{\cos A}{\cos B}. \] 2. Write the difference as a single fraction by finding a common denominator \(\sin B \cos B\): \[ \frac{\sin A \cos B - \cos A \sin B}{\sin B\cos B}. \] 3. Recognize that the numerator \(\sin A \cos B - \cos A \sin B\) is the sine of a difference: \[ \sin A \cos B - \cos A \sin B = \sin (A - B). \] 4. Thus, the expression becomes: \[ \frac{\sin (A - B)}{\sin B\cos B}. \] 5. Recall the double-angle formula for sine: \[ \sin 2B = 2\sin B \cos B. \] 6. Solve for \(\sin B\cos B\): \[ \sin B\cos B = \frac{\sin 2B}{2}. \] 7. Substitute \(\sin B\cos B = \frac{\sin 2B}{2}\) into the expression: \[ \frac{\sin (A - B)}{\frac{\sin 2B}{2}} = \frac{2\sin (A - B)}{\sin 2B}. \] 8. This shows that: \[ \frac{\sin A}{\sin B} - \frac{\cos A}{\cos B} = \frac{2\sin (A-B)}{\sin 2B}. \]