$T(n) = T\left(\frac{n}{\lg n}\right) + 1$
$= T\left(\frac{n}{\lg n \lg (n/ \lg n)}\right) + 2$ (Going 1 level into recursion)
$= T\left(\frac{n}{ \lg ^2 n - \lg n \lg \lg n}\right) + 2$
$\dots$
Since the second term $(\lg n \lg \lg n)$ is asymptotically lower than $\lg^2 n$, we can ignore it and assume that the recurrence stops when the denominator is half of the numerator (when we get T(2)) which happens after $k$ steps and since the complexity at each step is 1, the total complexity will be $k$.
$\lg^k n = n/2$
$\lg n = (n/2)^{1/k}$
$\lg \lg n = (1/k) (\lg n -1)$
$k = (\lg n -1 )/\lg \lg n$
So, time complexity = $\Theta \left(\frac{\lg n}{\lg \lg n}\right)$