\[
\tan\left(\dfrac{B-C}{2}\right)=\dfrac{b-c}{b+c}\cot{\left(\dfrac{A}{2}\right)}
\]
Com que aquesta relació es satisfà en un triangle (\(A+B+C=\pi\))
\[B-C=(\pi -A-C)-C=\pi-A-2C\]
i llavors
\[\tan\left(\dfrac{B-C}{2}\right)=\tan\left(\dfrac{\pi}{2}-\dfrac{A}{2}-C\right)=\dfrac{1}{\tan\left(\dfrac{A}{2}+C\right)}\]
degut a la propietat de la tangent
\[\tan\left(\dfrac{\pi}{2}-\theta\right)=\dfrac{1}{\tan\theta}\]
anem a veure que podem fer
\[
\begin{array}{rl}
\tan\left(\dfrac{B-C}{2}\right)=\dfrac{1}{\tan\left(\dfrac{A}{2}+C\right)}
= \dfrac{\cos\left(\dfrac{A}{2}+C\right)}{\sin\left(\dfrac{A}{2}+C\right)}
& = \dfrac{
\cos\left(\dfrac{A}{2}\right)\cos C-\sin\left(\dfrac{A}{2}\right)\sin C
}{
\sin\left(\dfrac{A}{2}\right)\cos C+\cos\left(\dfrac{A}{2}\right)\sin C
} \\
& = \dfrac{
\cos\left(\dfrac{A}{2}\right)\cos C-\sin\left(\dfrac{A}{2}\right)\dfrac{c}{a}\sin A
}{
\sin\left(\dfrac{A}{2}\right)\cos C+\cos\left(\dfrac{A}{2}\right)\dfrac{c}{a}\sin A
}
\end{array}
\]
on hem passat el \(\sin C\) en funció del \(\sin A\) utilitzant el Teorema del Sinus
\[
\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} \hspace{20pt} \sin C=\dfrac{c}{a}\sin A
\]
i ara podem passar aquest \(A\) a \(\dfrac{A}{2}\) utilizant la formula de l'angle doble
\[
\sin A = \sin 2\dfrac{A}{2} = 2\sin\dfrac{A}{2}\cos\dfrac{A}{2}
\]
i per tant aquell \(\sin C\) es transformaria en
\[
\sin C=\dfrac{c}{a}\sin A =\dfrac{c}{a}2\sin\left(\dfrac{A}{2}\right)\cos\left(\dfrac{A}{2}\right)
\]
introduit tot a la formula
\[
\begin{array}{rl}
\tan\left(\dfrac{B-C}{2}\right)
= \dfrac{
\cos\left(\dfrac{A}{2}\right)\cos C-\sin\left(\dfrac{A}{2}\right)\sin C
}{
\sin\left(\dfrac{A}{2}\right)\cos C+\cos\left(\dfrac{A}{2}\right)\sin C
}
& = \dfrac{
\cos\left(\dfrac{A}{2}\right)\cos C-\sin\left(\dfrac{A}{2}\right)\dfrac{c}{a}2\sin\left(\dfrac{A}{2}\right)\cos\left(\dfrac{A}{2}\right)
}{
\sin\left(\dfrac{A}{2}\right)\cos C+\cos\left(\dfrac{A}{2}\right)\dfrac{c}{a}2\sin\left(\dfrac{A}{2}\right)\cos\left(\dfrac{A}{2}\right)
} \\
& = \dfrac{
\cos\left(\dfrac{A}{2}\right)\cos C-2\dfrac{c}{a}\sin^2\left(\dfrac{A}{2}\right)\cos\left(\dfrac{A}{2}\right)
}{
\sin\left(\dfrac{A}{2}\right)\cos C+2\dfrac{c}{a}\cos^2\left(\dfrac{A}{2}\right)\sin\left(\dfrac{A}{2}\right)
}
\end{array}
\]
ara podem treure factor comú del cosinus al numerador i del sinus al denominador
\[
\begin{array}{rl}
\tan\left(\dfrac{B-C}{2}\right)
& = \dfrac{
\cos\left(\dfrac{A}{2}\right)\cos C-2\dfrac{c}{a}\sin^2\left(\dfrac{A}{2}\right)\cos\left(\dfrac{A}{2}\right)
}{
\sin\left(\dfrac{A}{2}\right)\cos C+2\dfrac{c}{a}\cos^2\left(\dfrac{A}{2}\right)\sin\left(\dfrac{A}{2}\right)
} \\
& = \dfrac{
\cos\left(\dfrac{A}{2}\right)\left[\cos C-2\dfrac{c}{a}\sin^2\left(\dfrac{A}{2}\right)\right]
}{
\sin\left(\dfrac{A}{2}\right)\left[\cos C+2\dfrac{c}{a}\cos^2\left(\dfrac{A}{2}\right)\right]
} \\
& = \dfrac{\cos\left(\dfrac{A}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}\cdot
\dfrac{
\cos C-2\dfrac{c}{a}\sin^2\left(\dfrac{A}{2}\right)
}{
\cos C+2\dfrac{c}{a}\cos^2\left(\dfrac{A}{2}\right)
} \\
& = \dfrac{1}{\tan\left(\dfrac{A}{2}\right)}
\dfrac{
\cos C-\dfrac{c}{a}2\sin^2\left(\dfrac{A}{2}\right)
}{
\cos C+\dfrac{c}{a}2\cos^2\left(\dfrac{A}{2}\right)
}
\end{array}
\]
utilitzant les formules de l'angle meitat
\[
2\sin^2\left(\dfrac{A}{2}\right) = 1 - \cos A \hspace{40pt}
2\cos^2\left(\dfrac{A}{2}\right) = 1 + \cos A
\]
llavors queda en funció només de cosinus
\[
\begin{array}{rl}
\tan\left(\dfrac{B-C}{2}\right)
& = \dfrac{1}{\tan\left(\dfrac{A}{2}\right)}
\dfrac{
\cos C-\dfrac{c}{a}2\sin^2\left(\dfrac{A}{2}\right)
}{
\cos C+\dfrac{c}{a}2\cos^2\left(\dfrac{A}{2}\right)
} \\
& = \dfrac{1}{\tan\left(\dfrac{A}{2}\right)}
\dfrac{
\cos C-\dfrac{c}{a}\left(1 - \cos A\right)
}{
\cos C+\dfrac{c}{a}\left(1 + \cos A\right)
}
\end{array}
\]
Utilitzant el Teorema del Cosinus pels caso que ens interessen:
\[
\cos A= \dfrac{b^2+c^2-a^2}{2bc} \hspace{40pt} \cos C= \dfrac{a^2+b^2-c^2}{2ab}
\]
Així
\[
\begin{array}{rl}
\tan\left(\dfrac{B-C}{2}\right)
= \dfrac{1}{\tan\left(\dfrac{A}{2}\right)}
\dfrac{
\cos C-\dfrac{c}{a}\left(1 - \cos A\right)
}{
\cos C+\dfrac{c}{a}\left(1 + \cos A\right)
}
& = \dfrac{1}{\tan\left(\dfrac{A}{2}\right)}
\dfrac{
\dfrac{a^2+b^2-c^2}{2ab} - \dfrac{c}{a}\left(1 - \dfrac{b^2+c^2-a^2}{2bc}\right)
}{
\dfrac{a^2+b^2-c^2}{2ab} + \dfrac{c}{a}\left(1 + \dfrac{b^2+c^2-a^2}{2bc}\right)
} \\
& = \dfrac{1}{\tan\left(\dfrac{A}{2}\right)}
\dfrac{
\dfrac{a^2+b^2-c^2}{2ab} - \dfrac{c}{a} + \dfrac{b^2+c^2-a^2}{2ab}
}{
\dfrac{a^2+b^2-c^2}{2ab} + \dfrac{c}{a} + \dfrac{b^2+c^2-a^2}{2ab}
} \\
& = \dfrac{1}{\tan\left(\dfrac{A}{2}\right)}
\dfrac{
\dfrac{a^2+b^2-c^2}{2ab} - \dfrac{2bc}{2ba} + \dfrac{b^2+c^2-a^2}{2ab}
}{
\dfrac{a^2+b^2-c^2}{2ab} + \dfrac{2bc}{2ba} + \dfrac{b^2+c^2-a^2}{2ab}
} \\
& = \dfrac{1}{\tan\left(\dfrac{A}{2}\right)}
\dfrac{
a^2+b^2-c^2 - 2bc + b^2+c^2-a^2
}{
a^2+b^2-c^2 + 2bc + b^2+c^2-a^2
} \\
\end{array}
\]
on hi han moltes cancel~lacions i
\[
\begin{array}{rl}
\tan\left(\dfrac{B-C}{2}\right)
& = \cot\left(\dfrac{A}{2}\right)
\dfrac{
2b^2 - 2bc
}{
2b^2 + 2bc
} \\
& = \cot\left(\dfrac{A}{2}\right)
\dfrac{
2b(b - c)
}{
2b(b + c)
} \\
& = \dfrac{ b - c }{b + c} \cot\left(\dfrac{A}{2}\right)
\end{array}
\]
q.e.d.
\[
\boxed{
\tan\left(\dfrac{B-C}{2}\right)=\dfrac{b-c}{b+c}\cot{\left(\dfrac{A}{2}\right)}
}
\]