Question

This is my question
$\tan( \frac{b - c}{2} ) = \frac{b - c}{b + c} \cot( \frac{a}{2} )$
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Answer 1

Step-by-step explanation:

According to sine rule, we have,

$\frac{ \sin(B) }{b} = \frac{ \sin(C) }{c}$

$\implies\frac{ \sin(B) }{\sin(C)} = \frac{ b }{c}$

$\rarr\tt \red{Applying\:\:componendo\:\:and\:\:dividendo}$

$\implies\frac{ \sin(B) + \sin(C)}{ \sin(B) - \sin(C) } = \frac{ b + c}{b - c}$

$\implies\frac{ 2\sin \bigg( \dfrac{B + C}{2} \bigg)\cos \bigg( \dfrac{B - C}{2} \bigg)}{ 2 \cos\bigg( \dfrac{B + C}{2} \bigg)\sin \bigg( \dfrac{B - C}{2} \bigg)} = \frac{ b + c}{b - c}$

$\implies\frac{ \sin \bigg( \dfrac{\pi - A}{2} \bigg)\cos \bigg( \dfrac{B - C}{2} \bigg)}{ \cos\bigg( \dfrac{\pi - A}{2} \bigg)\sin \bigg( \dfrac{B - C}{2} \bigg)} = \frac{ b + c}{b - c}$

$\implies\frac{ \sin \bigg( \dfrac{\pi}{2} - \dfrac{ A}{2} \bigg)\cos \bigg( \dfrac{B - C}{2} \bigg)}{ \cos\bigg( \dfrac{\pi}{2} - \dfrac{ A}{2} \bigg)\sin \bigg( \dfrac{B - C}{2} \bigg)} = \frac{ b + c}{b - c}$

$\implies\frac{ \cos \bigg( \dfrac{ A}{2} \bigg)\cos \bigg( \dfrac{B - C}{2} \bigg)}{ \sin\bigg( \dfrac{ A}{2} \bigg)\sin \bigg( \dfrac{B - C}{2} \bigg)} = \frac{ b + c}{b - c}$

$\implies \frac{\cos\bigg( \dfrac{ A}{2} \bigg)}{\sin\bigg( \dfrac{ A}{2} \bigg)} .\frac{\cos \bigg( \dfrac{B - C}{2} \bigg)}{ \sin \bigg( \dfrac{B - C}{2} \bigg)} = \frac{ b + c}{b - c}$

$\implies \cot\bigg( \dfrac{ A}{2} \bigg) .\frac{1}{ \tan\bigg( \dfrac{B - C}{2} \bigg)} = \frac{ b + c}{b - c}$

$\implies \frac{b - c}{b + c} \cot\bigg( \dfrac{ A}{2} \bigg) = \tan\bigg( \dfrac{B - C}{2} \bigg)$

$\purple{ \tt \bold{\implies \tan\bigg( \dfrac{B - C}{2} \bigg) = \frac{b - c}{b + c} \cot\bigg( \dfrac{ A}{2} \bigg)}}$