In a triangle ABC,the value of Sigma(b+c) tan | vedclass

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(D) We use the Napier's analogy: $	an \left(\frac{B-C}{2}ight) = \frac{b-c}{b+c} \cot \frac{A}{2}$.Substituting this into the expression:$(b+c) 	a

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(D) We use the Napier's analogy: $	an \left(\frac{B-C}{2}ight) = \frac{b-c}{b+c} \cot \frac{A}{2}$.Substituting this into the expression:$(b+c) 	an \frac{A}{2} 	an \left(\frac{B-C}{2}ight) = (b+c) 	an \frac{A}{2} \left[ \frac{b-c}{b+c} \cot \frac{A}{2} ight]$.Since $	an \frac{A}{2} \cot \frac{A}{2} = 1$,the expression simplifies to $(b-c)$.Now,summing over the cyclic permutations:$\Sigma(b-c) = (b-c) + (c-a) + (a-b) = 0$.

In a $\triangle ABC$,the value of $\Sigma(b+c) \tan \frac{A}{2} \tan \left(\frac{B-C}{2}\right)$ is equal to

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