Evaluate cot left( frac{A + B}{2} right) cdot | vedclass

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(B) Using the Napier's Analogy (Tangent Rule) in a triangle $ABC$:$	an \left( \frac{A - B}{2} ight) = \frac{a - b}{a   b} \cot \left( \frac{C}{2} \

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$\frac{a - b}{a + b}$

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(B) Using the Napier's Analogy (Tangent Rule) in a triangle $ABC$:$	an \left( \frac{A - B}{2} ight) = \frac{a - b}{a   b} \cot \left( \frac{C}{2} ight)$Since $A   B   C = 180^{\circ}$,we have $\frac{C}{2} = 90^{\circ} - \frac{A   B}{2}$.Therefore,$\cot \left( \frac{C}{2} ight) = \cot \left( 90^{\circ} - \frac{A   B}{2} ight) = 	an \left( \frac{A   B}{2} ight)$.Substituting this into the formula:$	an \left( \frac{A - B}{2} ight) = \frac{a - b}{a   b} 	an \left( \frac{A   B}{2} ight)$Multiplying both sides by $\cot \left( \frac{A   B}{2} ight)$:$\cot \left( \frac{A   B}{2} ight) \cdot 	an \left( \frac{A - B}{2} ight) = \frac{a - b}{a   b}$.

Evaluate $\cot \left( \frac{A + B}{2} \right) \cdot \tan \left( \frac{A - B}{2} \right)$ in terms of sides $a$ and $b$.

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