In a triangle ABC , with usual notations, $\cot \left(\frac{A+B}{2}\right) \cdot \tan \left(\frac{A-B}{2}\right)=$

In triangle $ABC$ where $A + B + C = \pi$, the expression $\cot \left(\frac{A+B}{2}\right) \cdot \tan \left(\frac{A-B}{2}\right)$ simplifies as follows.

Using the angle sum property, $\frac{A+B}{2} = \frac{\pi - C}{2} = \frac{\pi}{2} - \frac{C}{2}$.
By the complementary angle identity $\cot\left(\frac{\pi}{2} - x\right) = \tan x$, we obtain:
$\cot \left(\frac{A+B}{2}\right) = \tan \left(\frac{C}{2}\right)$.

Napier's Analogy gives $\tan \left(\frac{A-B}{2}\right) = \frac{a-b}{a+b} \cot \left(\frac{C}{2}\right)$.
Since $\cot \left(\frac{C}{2}\right) = \frac{1}{\tan \left(\frac{C}{2}\right)}$, this becomes:
$\tan \left(\frac{A-B}{2}\right) = \frac{a-b}{a+b} \cdot \frac{1}{\tan \left(\frac{C}{2}\right)}$.

Substituting both results into the original expression:
$\tan \left(\frac{C}{2}\right) \cdot \left( \frac{a-b}{a+b} \cdot \frac{1}{\tan \left(\frac{C}{2}\right)} \right) = \frac{a-b}{a+b}$.

Thus, the simplified expression is $\frac{a-b}{a+b}$.