1. **Problem statement:** Given a quadratic equation $ax^2+bx+c=0$ whose roots are $\cos \theta$ and $\sin \theta$, show that $$a^2-b^2+2ac=0.$$\n\n2. **Recall Vieta's formulas:** For roots $r_1$ and $r_2$ of $ax^2+bx+c=0$,\n$$r_1 + r_2 = -\frac{b}{a}, \quad r_1 r_2 = \frac{c}{a}.$$\n\n3. **Apply to the given roots:** Here, \n$$r_1 = \cos \theta, \quad r_2 = \sin \theta.$$\nThus, we have \n$$\cos \theta + \sin \theta = -\frac{b}{a}, \quad \cos \theta \sin \theta = \frac{c}{a}.$$\n\n4. **Square the sum of roots:**\n$$\left(\cos \theta + \sin \theta\right)^2 = \left(-\frac{b}{a}\right)^2 = \frac{b^2}{a^2}.$$\n\nExpanding the left side, \n$$\cos^2 \theta + 2 \cos \theta \sin \theta + \sin^2 \theta = \frac{b^2}{a^2}.$$\n\nSince $\cos^2 \theta + \sin^2 \theta = 1$,\n$$1 + 2 \cos \theta \sin \theta = \frac{b^2}{a^2}.$$\n\n5. **Substitute the product of roots:**\n$$1 + 2 \frac{c}{a} = \frac{b^2}{a^2}.$$\nMultiply both sides by $a^2$:\n$$a^2 + 2 a c = b^2.$$\n\n6. **Rearrange to the desired identity:**\n$$a^2 - b^2 + 2 a c = 0.$$\n\n**Conclusion:** We have shown that $$a^2-b^2+2ac=0$$ as required.