1. **State the problem:** Show that $$\frac{\cos 3a - \cos a}{\sin 3a + \sin a} = -\tan a$$. 2. **Recall formulas:** Use the sum-to-product identities: $$\cos A - \cos B = -2 \sin \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)$$ $$\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$$ 3. **Apply formulas:** Let $$A=3a$$ and $$B=a$$. Numerator: $$\cos 3a - \cos a = -2 \sin \left( \frac{3a + a}{2} \right) \sin \left( \frac{3a - a}{2} \right) = -2 \sin 2a \sin a$$ Denominator: $$\sin 3a + \sin a = 2 \sin \left( \frac{3a + a}{2} \right) \cos \left( \frac{3a - a}{2} \right) = 2 \sin 2a \cos a$$ 4. **Rewrite the fraction:** $$\frac{\cos 3a - \cos a}{\sin 3a + \sin a} = \frac{-2 \sin 2a \sin a}{2 \sin 2a \cos a}$$ 5. **Cancel common factors:** $$= \frac{\cancel{-2} \cancel{\sin 2a} \sin a}{\cancel{2} \cancel{\sin 2a} \cos a} = \frac{-\sin a}{\cos a}$$ 6. **Recognize the result:** $$\frac{-\sin a}{\cos a} = -\tan a$$ **Final answer:** $$\frac{\cos 3a - \cos a}{\sin 3a + \sin a} = -\tan a$$