7. Given a hyperbola with the two coordinate axes as axes of symmetry, the foci on the $y$-axis, the real axis length is $2 \sin \theta, \theta \in\left[\frac{\pi}{4}, \frac{\pi}{3}\right]$, and the shortest distance from any point $P(x, y)$ on the hyperbola to the point $M(1,0)$ is $\frac{1}{\sin \theta}$, then the range of the eccentricity of the hyperbola is . $\qquad$
7. $\left(1, \frac{2 \sqrt{21}}{7}\right]$ Detailed explanation: Let the equation of the hyperbola be $\frac{y^{2}}{\sin ^{2} \theta}-\frac{x^{2}}{b^{2}}=1$, then $|P M|^{2}=(x-1)^{2}+y^{2}=(x-1)^{2}+$ $\sin ^{2} \theta\left(1+\frac{x^{2}}{b^{2}}\right)=\left(1+\frac{\sin ^{2} \theta}{b^{2}}\right) x^{2}-2 x+1+\sin ^{2} \theta$, since $x \in \mathbf{R}$, hence $|P M|_{\text {min }}^{2}=1+\sin ^{2} \theta-\frac{b^{2}}{b^{2}+\sin ^{2} \theta}$, and because $|P M|_{\text {min }}^{2}=\frac{1}{\sin ^{2} \theta}$, it follows that $b^{2}=\frac{\sin ^{6} \theta+\sin ^{4} \theta-\sin ^{2} \theta}{1-\sin ^{4} \theta}$, and $b^{2}>0$ as well as $\theta \in\left[\frac{\pi}{4}, \frac{\pi}{3}\right]$. Solving the inequality yields $\frac{\sqrt{5}-1}{2}<\sin ^{2} \theta$ $\leqslant \frac{3}{4}$, and since $e^{2}=\left(\frac{c}{a}\right)^{2}=\frac{b^{2}+\sin ^{2} \theta}{\sin ^{2} \theta}=\frac{\sin ^{2} \theta}{1-\sin ^{4} \theta}=\frac{1}{\frac{1}{\sin ^{2} \theta}-\sin ^{2} \theta}$, let $t=\sin ^{2} \theta$, then $e^{2}=\frac{1}{\frac{1}{t}-t}$, as $f(t)=$ $\frac{1}{1}-t$ is an increasing function on $\left(\frac{\sqrt{5}-1}{2}, \frac{3}{4}\right]$, hence $1<e^{2} \leqslant \frac{12}{7}, 1<e \leqslant \frac{2 \sqrt{21}}{7}$.