Find the least positive integer $n$ such that \[ \frac 1{\sin 45^\circ\sin 46^\circ}+\frac 1{\sin 47^\circ\sin 48^\circ}+\cdots+\frac 1{\sin 133^\circ\sin 134^\circ}=\frac 1{\sin n^\circ}. \]
1. We start with the given equation:
   \[
   \frac{1}{\sin 45^\circ \sin 46^\circ} + \frac{1}{\sin 47^\circ \sin 48^\circ} + \cdots + \frac{1}{\sin 133^\circ \sin 134^\circ} = \frac{1}{\sin n^\circ}
   \]

2. Notice that \(\sin(180^\circ - x) = \sin x\). Therefore, we can rewrite the sines of angles greater than \(90^\circ\) in terms of their supplementary angles:
   \[
   \sin 134^\circ = \sin 46^\circ, \sin 133^\circ = \sin 47^\circ, \ldots, \sin 91^\circ = \sin 89^\circ
   \]

3. This allows us to pair the terms symmetrically around \(90^\circ\):
   \[
   \frac{1}{\sin 45^\circ \sin 46^\circ} + \frac{1}{\sin 47^\circ \sin 48^\circ} + \cdots + \frac{1}{\sin 89^\circ \sin 91^\circ} + \cdots + \frac{1}{\sin 46^\circ \sin 45^\circ}
   \]

4. Using the identity \(\sin x = \cos (90^\circ - x)\), we can rewrite the terms:
   \[
   \frac{1}{\cos 45^\circ \cos 44^\circ} + \frac{1}{\cos 43^\circ \cos 42^\circ} + \cdots + \frac{1}{\cos 1^\circ \cos 0^\circ} + \frac{1}{\cos 1^\circ \cos 2^\circ} + \cdots + \frac{1}{\cos 44^\circ \cos 45^\circ}
   \]

5. Since \(\cos x\) is an even function, \(\cos(-x) = \cos x\), we can simplify the expression:
   \[
   \frac{1}{\cos 0^\circ \cos 1^\circ} + \frac{1}{\cos 1^\circ \cos 2^\circ} + \cdots + \frac{1}{\cos 44^\circ \cos 45^\circ}
   \]

6. Rewriting in terms of secant:
   \[
   \sec 0^\circ \sec 1^\circ + \sec 1^\circ \sec 2^\circ + \sec 2^\circ \sec 3^\circ + \cdots + \sec 44^\circ \sec 45^\circ
   \]

7. We approximate \(\sec(x) \sec(x + 1^\circ)\) by \(\sec^2 x\) since the error is small:
   \[
   \sec^2 0^\circ + \sec^2 1^\circ + \sec^2 2^\circ + \cdots + \sec^2 44^\circ
   \]

8. The sum of \(\sec^2 x\) can be approximated by the integral of \(\sec^2 x\) over the interval \([0, \frac{\pi}{4}]\):
   \[
   \sum_{k=0}^{44} \sec^2 k^\circ \approx \frac{180}{\pi} \int_{0}^{\frac{\pi}{4}} \sec^2 x \, dx
   \]

9. Evaluating the integral:
   \[
   \int_{0}^{\frac{\pi}{4}} \sec^2 x \, dx = \left. \tan x \right|_{0}^{\frac{\pi}{4}} = \tan \frac{\pi}{4} - \tan 0 = 1 - 0 = 1
   \]

10. Therefore:
    \[
    \frac{180}{\pi} \cdot 1 = \frac{180}{\pi}
    \]

11. We equate this to \(\frac{1}{\sin n^\circ}\):
    \[
    \frac{180}{\pi} = \frac{1}{\sin n^\circ}
    \]

12. Solving for \(\sin n^\circ\):
    \[
    \sin n^\circ = \frac{\pi}{180}
    \]

13. Near the origin, \(\sin x \approx x\) in radians. Converting \(\frac{\pi}{180}\) to degrees:
    \[
    n \approx 1^\circ
    \]

The final answer is \(\boxed{1}\).