(D) We know the Napier's analogy: $\tan \left(\frac{B-C}{2}\right) = \frac{b-c}{b+c} \cot \frac{A}{2}$.Rearranging this,we get $(b+c) \tan \frac{A}{2}

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(D) We know the Napier's analogy: $\tan \left(\frac{B-C}{2}\right) = \frac{b-c}{b+c} \cot \frac{A}{2}$.Rearranging this,we get $(b+c) \tan \frac{A}{2} \tan \left(\frac{B-C}{2}\right) = (b-c)$.Now,applying the summation $\Sigma$ over the cyclic terms:$\Sigma(b+c) \tan \frac{A}{2} \tan \left(\frac{B-C}{2}\right) = (b-c) + (c-a) + (a-b)$.Summing these terms: $b - c + c - a + a - b = 0$.

Class 11 Mathematics Trigonometrical Equations Mix Examples-Trigonometrical Equations and Inequations, Properties of Triangles, Height and Distance

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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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Mix Examples-Trigonometrical Equations and Inequations, Properties of Triangles, Height and Distance

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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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