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Variable Separable Method

In triangle ...

Question

In triangle $A B C,$ prove that :
$tan \frac{B - C}{2} = \frac{b - c}{b + c} cot \frac{A}{2}$

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Solution

$\begin{matrix} I n Δ A B C \frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} = 2 R (l e t) \frac{b - c}{b + c} = \frac{2 R sin B - 2 R sin C}{2 R sin B + 2 R sin C} = \frac{sin B - sin C}{sin B + sin C} = \frac{2 cos \frac{B + C}{2} . sin \frac{B - C}{2}}{2 sin \frac{B + C}{2} . cos \frac{B - C}{2}} = cot \frac{B + C}{2} . tan \frac{B - C}{2} = tan \frac{A}{2} . tan \frac{B - C}{2} ∴ tan \frac{B - C}{2} = \frac{b - c}{b + c} cot \frac{A}{2} \end{matrix}$

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