1. **Stating the problem:** We are given the equation $m^2 - n^2 = 4\sqrt{mn}$ and the expression $m = \tan A + \sin A$. We need to prove that $n = \tan A - \sin A$. 2. **Recall the difference of squares formula:** $$m^2 - n^2 = (m - n)(m + n)$$ 3. **Rewrite the given equation using this formula:** $$(m - n)(m + n) = 4\sqrt{mn}$$ 4. **Substitute $m = \tan A + \sin A$ and assume $n = \tan A - \sin A$ to verify:** Calculate $m - n$: $$m - n = (\tan A + \sin A) - (\tan A - \sin A) = 2\sin A$$ Calculate $m + n$: $$m + n = (\tan A + \sin A) + (\tan A - \sin A) = 2\tan A$$ 5. **Calculate the product:** $$(m - n)(m + n) = 2\sin A \times 2\tan A = 4 \sin A \tan A$$ 6. **Calculate $4\sqrt{mn}$:** First find $mn$: $$mn = (\tan A + \sin A)(\tan A - \sin A) = \tan^2 A - \sin^2 A$$ 7. **Simplify $\tan^2 A - \sin^2 A$ using trigonometric identities:** Recall $\tan A = \frac{\sin A}{\cos A}$, so $$\tan^2 A = \frac{\sin^2 A}{\cos^2 A}$$ Thus, $$\tan^2 A - \sin^2 A = \frac{\sin^2 A}{\cos^2 A} - \sin^2 A = \sin^2 A \left(\frac{1}{\cos^2 A} - 1\right) = \sin^2 A \frac{1 - \cos^2 A}{\cos^2 A}$$ Since $1 - \cos^2 A = \sin^2 A$, we get $$mn = \sin^2 A \frac{\sin^2 A}{\cos^2 A} = \frac{\sin^4 A}{\cos^2 A}$$ 8. **Calculate $4\sqrt{mn}$:** $$4\sqrt{mn} = 4 \sqrt{\frac{\sin^4 A}{\cos^2 A}} = 4 \frac{\sin^2 A}{\cos A} = 4 \sin A \times \frac{\sin A}{\cos A} = 4 \sin A \tan A$$ 9. **Compare both sides:** $$(m - n)(m + n) = 4 \sin A \tan A = 4 \sqrt{mn}$$ This matches the given equation, confirming that $n = \tan A - \sin A$ is true. **Final answer:** $$\boxed{n = \tan A - \sin A}$$