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In a $\triangle A B C, b=\sqrt{3}+1, c=\sqrt{3}-1, \angle A=60^{\circ}$ then the value of $\tan \left(\frac{B-C}{2}\right)$ is
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The correct answer is:
$1$
$\tan \left(\frac{B-C}{2}\right)=\left(\frac{b-c}{b+c}\right) \cot \left(\frac{A}{2}\right)$
putting the value of $\mathrm{b}, \mathrm{c}$ and $\angle \mathrm{A}$
$\tan \left(\frac{B-C}{2}\right)=\frac{(\sqrt{3}+1)-(\sqrt{3}-1)}{(\sqrt{3}+1)+(\sqrt{3}-1)} \cot \left(30^{\circ}\right)$
$\Rightarrow \quad \tan \left(\frac{B-C}{2}\right)=1$
putting the value of $\mathrm{b}, \mathrm{c}$ and $\angle \mathrm{A}$
$\tan \left(\frac{B-C}{2}\right)=\frac{(\sqrt{3}+1)-(\sqrt{3}-1)}{(\sqrt{3}+1)+(\sqrt{3}-1)} \cot \left(30^{\circ}\right)$
$\Rightarrow \quad \tan \left(\frac{B-C}{2}\right)=1$
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