In ∆ABC, tan\((\frac{B-C}{2})\) =(A) \(\frac{b-c}{b+c}\)cot\(\frac{A}{2}\)(B) \(\frac{b+c}{b-c}\)cot\(\frac{A}{2}\)(C) \(\frac{b^2-c^2}{b-c}\)cot\(\frac{A}{2}\)(D) \(\frac{b+c}{b-c}\)cot2\(\frac{A}{2}\)

In ∆ABC, tan\((\frac{B-C}{2})\) =(A) \(\frac{b-c}{b+c}\)cot\(\frac

1.	In ∆ABC, tan\((\frac{B-C}{2})\) =(A) \(\frac{b-c}{b+c}\)cot\(\frac{A}{2}\)(B) \(\frac{b+c}{b-c}\)cot\(\frac{A}{2}\)(C) \(\frac{b^2-c^2}{b-c}\)cot\(\frac{A}{2}\)(D) \(\frac{b+c}{b-c}\)cot2\(\frac{A}{2}\)
Answer» Correct option is: (A) \(\frac{b-c}{b+c}\)cot\(\frac{A}{2}\)

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In ∆ABC, tan\((\frac{B-C}{2})\) =(A) \(\frac{b-c}{b+c}\)cot\(\frac{A}{2}\)(B) \(\frac{b+c}{b-c}\)cot\(\frac{A}{2}\)(C) \(\frac{b^2-c^2}{b-c}\)cot\(\frac{A}{2}\)(D) \(\frac{b+c}{b-c}\)cot2\(\frac{A}{2}\)

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