Line $l$ tangents unit circle $S$ in point $P$. Point $A$ and circle $S$ are on the same side of $l$, and the distance from $A$ to $l$ is $h$ ($h > 2$). Two tangents of circle $S$ are drawn from $A$, and intersect line $l$ at points $B$ and $C$ respectively. 

Find the value of $PB \cdot PC$.
1. **Define the problem setup:**
   - Let \( S \) be the unit circle with radius \( r = 1 \).
   - Let \( l \) be the tangent line to \( S \) at point \( P \).
   - Let \( A \) be a point such that the distance from \( A \) to \( l \) is \( h \) (where \( h > 2 \)).
   - From \( A \), two tangents to the circle \( S \) are drawn, intersecting \( l \) at points \( B \) and \( C \).

2. **Determine the relationship between the tangents and the circle:**
   - The tangents from a point outside a circle to the circle are equal in length. Therefore, \( AB = AC \).

3. **Use the power of a point theorem:**
   - The power of point \( A \) with respect to the circle \( S \) is given by:
     \[
     PA^2 - r^2 = PB \cdot PC
     \]
   - Since \( r = 1 \), this simplifies to:
     \[
     PA^2 - 1 = PB \cdot PC
     \]

4. **Calculate \( PA \):**
   - The distance from \( A \) to the line \( l \) is \( h \). Since \( l \) is tangent to the circle at \( P \), the distance from \( A \) to \( P \) (along the perpendicular from \( A \) to \( l \)) is \( h \).
   - Therefore, \( PA = h \).

5. **Substitute \( PA \) into the power of a point formula:**
   \[
   h^2 - 1 = PB \cdot PC
   \]

6. **Simplify the expression:**
   \[
   PB \cdot PC = h^2 - 1
   \]

The final answer is \( \boxed{h^2 - 1} \).