Step 1: Use the triangle angle sum property.
In triangle $ABC$, $A+B+C = \pi$, so $\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi}{2}$. Let $p=\frac{A}{2}$, $q=\frac{B}{2}$, $r=\frac{C}{2}$. Then $p+q+r=\frac{\pi}{2}$.
Step 2: Use the identity for $p+q+r=\frac{\pi}{2}$.
If $p+q+r = \frac{\pi}{2}$, then $p+q = \frac{\pi}{2}-r$, so $\tan(p+q) = \cot r = \frac{1}{\tan r}$.
Step 3: Expand $\tan(p+q)$.
\[ \frac{\tan p+\tan q}{1-\tan p\tan q} = \frac{1}{\tan r} \] Cross multiply: $\tan r(\tan p+\tan q) = 1-\tan p\tan q$, giving: $\tan p\tan q + \tan q\tan r + \tan r\tan p = 1$.
Step 4: Translate back to original notation.
\[ \tan\frac{A}{2}\tan\frac{B}{2} + \tan\frac{B}{2}\tan\frac{C}{2} + \tan\frac{C}{2}\tan\frac{A}{2} = 1 \]
Step 5: Verify with an equilateral triangle.
For $A=B=C=60°$: each $\tan(30°)=\frac{1}{\sqrt{3}}$. Sum of products: $3 \times \frac{1}{3} = 1$. Confirmed.
Step 6: State the answer.
\[ \boxed{1} \]