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Using the Napier's Analogies(Laws of tangents)  (Source: Law of tangents):   $tan\left ( \frac{B-C}{2} \right )=\frac{b-c}{b+c}cot\left ( \frac{A}{2} \right )$

$\therefore$  $tan\left ( \frac{B-C}{2} \right )\times tan\left ( \frac{A}{2} \right )=\frac{b-c}{b+c}\times cot\left ( \frac{A}{2} \right )\times tan\left ( \frac{A}{2} \right )=\frac{b-c}{b+c}$

Option B is correct.

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