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In a $\triangle A B C, \Sigma(b+c) \tan \frac{A}{2} \tan \left(\frac{B-C}{2}\right)$ is equal to
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$\begin{aligned} & \Sigma(b+c) \tan \frac{A}{2} \tan \left(\frac{B-C}{2}\right) \\ & \therefore(b+c) \tan \frac{A}{2} \tan \left(\frac{B-C}{2}\right) \\ & =(b+c) \cdot \frac{(b-c)}{(b+c)} \cot \frac{A}{2} \tan \frac{A}{2} \\ & =b-c \\ & \therefore \quad \Sigma(b+c) \tan \frac{B-C}{2} \tan \frac{A}{2} \\ & =b-c+c-a+a-b=0 \\ & \end{aligned}$
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