1. Stating the problem: Given the equation $m \sin\theta = n \sin(\theta + 2)$, find the value of $\frac{m+n}{m-n} \tan \theta$. 2. Use the sine addition formula for $\sin(\theta + 2)$: $$\sin(\theta + 2) = \sin\theta \cos 2 + \cos\theta \sin 2$$ 3. Substitute into the original equation: $$m \sin\theta = n (\sin\theta \cos 2 + \cos\theta \sin 2)$$ 4. Rearrange terms: $$m \sin\theta = n \sin\theta \cos 2 + n \cos\theta \sin 2$$ 5. Group terms containing $\sin\theta$ and $\cos\theta$: $$m \sin\theta - n \sin\theta \cos 2 = n \cos\theta \sin 2$$ 6. Factor out $\sin\theta$ on the left: $$\sin\theta (m - n \cos 2) = n \cos\theta \sin 2$$ 7. Divide both sides by $\cos\theta (m - n \cos 2)$: $$\frac{\sin\theta}{\cos\theta} = \frac{n \sin 2}{m - n \cos 2}$$ 8. Recognize that $\frac{\sin\theta}{\cos\theta} = \tan\theta$: $$\tan\theta = \frac{n \sin 2}{m - n \cos 2}$$ 9. Find $\frac{m+n}{m-n} \tan\theta$ substituting for $\tan\theta$: $$\frac{m+n}{m-n} \tan\theta = \frac{m+n}{m-n} \cdot \frac{n \sin 2}{m - n \cos 2}$$ 10. Simplify the denominator $m - n \cos 2$ by writing expression explicitly, or leave as is since no further simplification is indicated. Hence the final answer is: $$\frac{m+n}{m-n} \tan \theta = \frac{n (m+n) \sin 2}{(m-n)(m - n \cos 2)}$$