\( 12.3 \quad \) 12.3.1 \( \quad \) Prove that, for angles \( A \) and \( B \) \[ \frac{\sin A}{\sin B}-\frac{\cos A}{\cos B}=\frac{2 \sin (A-B)}{\sin 2 B} \] 12.3.2 Hence, or otherwise, without using a calculator, show that: (a) \( \frac{\sin 5 B}{\sin B}-\frac{\cos 5 B}{\cos B}=4 \cos 2 B \) (b) \( \quad \frac{1}{\sin 18^{\circ}}=4 \cos 36^{\circ} \)

**Step 1. Prove that** \[ \frac{\sin A}{\sin B} - \frac{\cos A}{\cos B} = \frac{2\sin (A - B)}{\sin 2B}. \] 1. Write the left-hand side (LHS) with a common denominator: \[ \frac{\sin A}{\sin B} - \frac{\cos A}{\cos B} = \frac{\sin A \cos B - \cos A \sin B}{\sin B\,\cos B}. \] 2. Recognize that the numerator is the sine of a difference: \[ \sin A \cos B - \cos A \sin B = \sin (A - B). \] Thus, the expression becomes: \[ \frac{\sin (A - B)}{\sin B\,\cos B}. \] 3. Use the double-angle formula for sine: \[ \sin 2B = 2\sin B\,\cos B \quad \Longrightarrow \quad \sin B\,\cos B = \frac{1}{2}\sin 2B. \] 4. Substitute into the expression: \[ \frac{\sin (A - B)}{\tfrac{1}{2}\sin 2B} = \frac{2\sin (A - B)}{\sin 2B}. \] Thus, the identity is proved. --- **Step 2. (a) Show that** \[ \frac{\sin 5B}{\sin B} - \frac{\cos 5B}{\cos B} = 4\cos 2B. \] 1. Set \(A = 5B\) in the proved identity: \[ \frac{\sin 5B}{\sin B} - \frac{\cos 5B}{\cos B} = \frac{2\sin(5B - B)}{\sin 2B} = \frac{2\sin 4B}{\sin 2B}. \] 2. Express \(\sin 4B\) using the double-angle formula: \[ \sin 4B = 2\sin 2B\,\cos 2B. \] 3. Substitute into the expression: \[ \frac{2\cdot(2\sin 2B\,\cos 2B)}{\sin 2B} = \frac{4\sin 2B\,\cos 2B}{\sin 2B} = 4\cos 2B. \] Thus, the result holds. --- **Step 2. (b) Show that** \[ \frac{1}{\sin 18^\circ} = 4\cos 36^\circ. \] 1. Choose \(B = 18^\circ\). Then \(5B = 90^\circ\). 2. Substitute into the result of part (a): \[ \frac{\sin 5B}{\sin B} - \frac{\cos 5B}{\cos B} = 4\cos 2B. \] With \(5B = 90^\circ\) and \(2B = 36^\circ\), we have: \[ \frac{\sin 90^\circ}{\sin 18^\circ} - \frac{\cos 90^\circ}{\cos 18^\circ} = 4\cos 36^\circ. \] 3. Since \(\sin 90^\circ = 1\) and \(\cos 90^\circ = 0\), the left-hand side becomes: \[ \frac{1}{\sin 18^\circ} - 0 = \frac{1}{\sin 18^\circ}. \] Thus, \[ \frac{1}{\sin 18^\circ} = 4\cos 36^\circ, \] which completes the proof.