7. Given $5 \sin 2 \alpha=\sin 2^{\circ}$, then the value of $\frac{\tan \left(\alpha+1^{\circ}\right)}{\tan \left(\alpha-1^{\circ}\right)}$ is $\qquad$ -
The difficulty of this problem: Average for the first attempt
Mainly tests: Sum and difference angle formulas, trigonometric identities
Solution: $-\frac{2}{3}$
Because $\tan \left(\alpha+1^{\circ}\right)+\tan \left(\alpha-1^{\circ}\right)=\frac{\sin 2 \alpha}{\cos \left(\alpha+1^{\circ}\right) \cos \left(\alpha-1^{\circ}\right)}$
At the same time, $\tan \left(\alpha+1^{\circ}\right)-\tan \left(\alpha-1^{\circ}\right)=\frac{\sin 2^{\circ}}{\cos \left(\alpha+1^{\circ}\right) \cos \left(\alpha-1^{\circ}\right)}$
Therefore, $\frac{\tan \left(\alpha+1^{\circ}\right)+\tan \left(\alpha-1^{\circ}\right)}{\tan \left(\alpha+1^{\circ}\right)-\tan \left(\alpha-1^{\circ}\right)}=\frac{\sin 2 \alpha}{\sin 2^{\circ}}=\frac{1}{5}$
So $\frac{\tan \left(\alpha+1^{\circ}\right)}{\tan \left(\alpha-1^{\circ}\right)}=-\frac{3}{2}$